ELECTRICAL  ENGINEERING  TEXTS 


PRINCIPLES 

OF 

ALTERNATING  CURRENTS 


ELECTRICAL  ENGINEERING 
TEXTS 

A  series  of  textbooks  outlined  by  a  com- 
mittee of  well-known  electrical  engineers 
of  which  Harry  E.  Clifford,  Gordon  Mc- 
Kay Professor  of  Electrical  Engineering, 
Harvard  University,  is  Chairman  and 
Consulting  Editor. 
Laws — 

ELECTRICAL  MEASUREMENTS 
Lawrence — 

PRINCIPLES  OF  ALTERNATING-CUR- 
RENT MACHINERY 
Lawrence — 

PRINCIPLES  OF  ALTERNATING  CUR- 
RENTS 
Langsdorf — 
PRINCIPLES    OF    DIRECT-CURRENT 

MACHINES 
Dawes — 

COURSE  IN  ELECTRICAL  ENGINEER- 
ING 

Vol.     I. — Direct  Currents 
Vol.  II. — Alternating  Currents 
Dawes — 

INDUSTRIAL    ELECTRICITY— PART   I 
Dawes — 
INDUSTRIAL   ELECTRICITY— PART  II 


ELECTRICAL    ENGINEERING    TEXTS 


PRINCIPLES 


OF 


ALTERNATING   CURRENTS 


BY 
RALPH  R.  LAWRENCE 


ASSOCIATE   PROFESSOR   OF   ELECTRICAL  ENGINEERING   OF   THE   MASSACHUSETTS 

INSTITUTE    OF    TECHNOLOGY;    MEMBER   OF   THE  AMERICAN 

INSTITUTE    OF   ELECTRICAL   ENGINEERS 


FIRST  EDITION 
FOURTH  IMPRESSION 


McGRAW-HILL  BOOK  COMPANY,  INC, 
NEW  YORK:  370  SEVENTH  AVENUE 

LONDON:  6  &  8  BOUVERIE  ST.,  E.  C.  4 
1922 


TK  •/  14-1 


COPYRIGHT,  1922,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY,  INC. 

PRINTED  I N   THE   UNITED   STATES   OF   AMERICA 


THE  MAPLE  PRESS  -  YORK  PA 


PREFACE 

This  book  has  been  developed  from  notes  on  Alternating  Cur- 
rents used  for  several  years  at  the  Massachusetts  Institute  of 
Technology  with  the  junior  students  in  Electrical  Engineering. 
The  portions  of  the  notes  dealing  with  single-phase  currents  were 
originally  written  by  Professor  H.  E.  Clifford,  Gordon  McKay 
Professor  of  Electrical  Engineering  at  Harvard  University,  who 
was  formerly  Professor  of  Electrical  Engineering  at  the  Massa- 
chusetts Institute  of  Technology.  The  general  arrangement  and 
much  of  the  material  of  these  portions  of  the  book  are  sub- 
stantially in  the  same  form  as  originally  written. 

No  attempt  has  been  made  to  include  problems  other  than 
those  used  to  illustrate  the  principles  discussed,  as  two  sets  of 
problems  on  alternating  currents,  much  more  comprehensive 
than  could  have  been  incorporated  in  the  book,  were  already 
published  by  Professor  W.  V.  Lyon  under  the  titles  "  Problems 
in  Electrical  Engineering"  and  "Problems  in  Alternating  Cur- 
rent Machinery." 

The  author  wishes  to  express  his  indebtedness  to  Professor 
H.  E.  Clifford  for  the  care  with  which  he  edited  the  manuscript 
and  read  the  proof.  The  author  also  wishes  to  thank  Professor 
W.  V.  Lyon  for  his  many  suggestions. 

RALPH  R.  LAWRENCE. 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 
CAMBRIDGE,  January,  1922. 


640709 


NOTATION 

In  general  the  notation  recommended  by  the  American  Institute  of  Elec- 
trical Engineers  has  been  used.  When  the  significance  of  the  letters  differs 
from  that  given  in  the  following  table,  it  is  so  stated  in  the  text.  A  line 
over  a  letter  indicates  that  it  is  a  vector  or  complex  quantity. 

b  =  Susceptance. 

C  =  Capacitance. 

D  =  Distance. 

d  =  Distance. 

E  =  Root-mean-square  value  of  a  voltage  or  an  electromotive  force. 
Em  =  Maximum  value  of  a  voltage  or  an  electromotive  force. 

e  =  Instantaneous  value  of  a  voltage  or  an  electromotive  force. 

/  =  Frequency  in  cycles  per  second. 

f  =  Function. 

g  =  Conductance. 

h  =  Height  above    the    earth  of  the  conductor  of  a  transmission  line. 
3C  =  Field  intensity. 

/  =  Root-mean-square  value  of  a  current. 
Im  =  Maximum  value  of  a  current. 

i  =  Instantaneous  value  of  a  current. 

j  =  Operator  which  produces  a  counter-clockwise  rotation  of  ninety 
degrees. 

k  =  Constant. 

L  =  Coefficient  of  self-induction,  or  self-inductance,  or  a  length. 
M  =  Coefficient  of  mutual-induction,  or  mutual  inductance. 

m  =  Integer. 

N  =  Number  of  turns. 

n  =  Integer. 

O  =  Neutral  point. 

P  =  Average  power. 

p  =  Instantaneous  power. 
p.f.  =  Power-factor. 

Q  =  Root-mean-square  value  of  a  charge  or  steady  value  of  a  charge. 
Qm  =  Maximum  value  of  a  charge. 

q  =  Instantaneous  value  of  a  charge. 
(R    =  Magnetic  reluctance. 

r  =  Resistance. 

S  =  Coefficient  of  leakage  induction,  or  leakage-inductance. 

T  =  j  =  Time  in  seconds  of  a  complete  cycle. 

vii 


viii  NOTATION 

t  =  Time  in  seconds. 

V  =  Root-mean-square  value  of  a  voltage  or  an  electromotive  force. 
Vm  =  Maximum  value  of  a  voltage  or  an  electromotive  force. 

v  =  Instantaneous  value  of  a  voltage  or  an  electromotive  force. 
W  =  Energy. 
x  =  Reactance. 
y  =  Admittance. 
Z  =  Number  of  inductors. 
z  =  Impedance, 
a  =  Phase  angle. 
/3  =  Phase  angle. 
(B  =  Flux  density. 

€  =  2.718  =  Base  of  Napierian  logarithms. 
0  =  Phase  angle. 
X  =  Wave  length. 
M  =  Permeability, 
co  =  2irf  =  Angular  velocity. 
2  =  Summation. 
v  =  Flux. 


CONTENTS 

CHAPTER  1 

PAGE 

ALGEBRA  OF  VECTORS  AND  OF  COMPLEX  QUANTITIES  USED  IN  ELECTRI- 
CAL ENGINEERING 1 

Quantities  Involved  in  the  Solution  of  Problems  in  Alternating 
Currents — Types  of  Vectors  met  in  Electrical  Engineering  Prob- 
lems— Solution  of  Problems  Involving  Vectors  by  the  Method  of 
Trigonometry — Vector  Algebra — Method  of  Complex  Quantities — 
Representation  of  Vectors  by  the  Use  of  the  Operator  3 — Real  and 
Imaginary  Components  of  a  Vector  and  Real  and  Imaginary  Axes 
— Operator  (cos  a  ±  j  sin  a) — Operator  which  Rotates  the 
Reference  Axes  through  an  Angle  a — Successive  Application  of 
Rotating  Operators,  Powers  and  Roots  of  Operators,  Reciprocal  of 
an  Operator — Reciprocal  of  the  Operator  (cos  a.  ±  j  sin  a) —  Oper- 
ators which  Produce  Uniform  Angular  Rotation — Solution  of  Vector 
Equations  when  the  Vectors  and  Complex  Quantities  Involved  are 
Expressed  in  the  Complex  Form,  i.e.,  in  the  Form  a  +  jb — Expo- 
nential Operator  e  -  j° — Exponential  Operator  which  Produces 
Uniform  Rotation — Polar  Form  of  Operator — Square  Root,  Product 
and  Ratio  of  Vectors  or  Complex  Quantities  by  the  Use  of  the 
Exponential  Operator — Complex  Quantity — The  nth  Root  and  the 
nth  Power  of  Vectors  and  Complex  Quantities — Logarithm  of  a 
Complex  Quantity  or  a  Vector — Representation  of  an  Oscillating 
Vector,  whose  Magnitude  varies  Sinusoidally  with  Time,  by  the 
use  of  Two  Exponential  Rotating  Operators. 

CHAPTER  II 

ALTERNATING  CURRENTS '  .     27 

Direct  Current  or  Voltage — Pulsating  Current  or  Voltage — Con- 
tinuous Current  or  Voltage — Alternating  Current  or  Voltage — 
Oscillating  Current — Instantaneous  Value — Cycle — Periodic  Time 
or  Period — Frequency — Wave  Shape  or  Wave  Form — Simple  Har- 
monic Current  or  Voltage — Alternating  Current  Calculations 
Based  on  Sine  Waves — Phase — Generation  of  an  Alternating  Elec- 
tromotive Force — Strength  of  Current — Ampere  Value  of  an  Alter- 
nating Current — Volt  Value  of  an  Alternating  Voltage — Advantages 
of  Defining  the  Strength  of  Alternating  Currents  by  the  Square 
Root  of  their  Average  Square  Value — Relation  between  the  Root- 
mean-square  and  Average  Values  for  Simple  Harmonic  or  Sinu- 
soidal Currents — Form-factor — Measurement  of  the  Effective  or 
Root-mean-square  Value  of  a  Current  or  Voltage — Vector  Repre- 
sentation of  Simple  Harmonic  Currents  and  Voltages. 

ix 


X  CONTENTS 

CHAPTER  III 

PAGE 

POWER  WHEN  CURRENT  AND  VOLTAGE  ARE  SINUSOIDAL 48 

Power  Absorbed  and  Delivered  by  a  Circuit — Instantaneous  Power 
— Average  Power — Power  when  the  Voltage  and  Current  Waves 
are  both  Sinusoidal — Volt-amperes,  Apparent  or  Virtual  Power — 
Power-factor — Reactive-factor — Measurement  of  Average  Power — 
Measurement  of  Power-factor — Energy  and  Wattless  or  Quadra- 
ture Components  of  Current — Energy  and  Wattless  or  Quadrature 
Components  of  Voltage — Measurement  of  Reactive  Power,  i.e., 
Reactive  Volt-amperes — Vectors  Representing  Voltage  Rise  and 
Voltage  Fall — Expression  for  Power  when  the  Voltage  and  Current 
are  in  Complex. 

CHAPTER  IV 

NON-SINUSOIDAL  WAVES 69 

Wave  form  of  Alternators — Representation  of  a  Non-sinusoidal 
Current  or  Voltage  Wave  by  a  Fourier  Series — Effect  of  Even 
Harmonics  on  Wave  Form — Waves  which  have  the  Halves  of  the 
Positive  and  Negative  Loop  Symmetrical — Changing  the  Reference 
Point  from  which  Angles  and  Time  are  Measured  in  a  Complex 
Wave — Fourier  Series  for  Rectangular  and  Triangular  Waves — 
Measurement  of  Current,  Voltage  and  Power  when  the  Wave 
Form  is  not  Sinusoidal — Determination  of  Wave  Form — Effective 
Value  of  a  Non-sinusoidal  Electromotive  Force  or  Current — Power 
when  the  Electromotive  Force  and  Current  are  Non-sinusoidal 
Waves — Power-factor  when  the  Current  and  Voltage  are  not  Sinu- 
soidal— Effective  Value  of  an  Alternating  Electromotive  Force  or 
Current  from  a  Polar  Plot — Analysis  of  a  Non-sinusoidal  Wave  and 
the  Determination  of  Its  Fundamental  and  Harmonics — Fischer- 
Hinnen  Method  of  Analysing  a  Periodic  Wave  into  Com- 
ponents of  Its  Fouries  Series — Form  Factor — Amplitude,  Crest  or 
Peak  Factor — Deviation  Factor — Equivalent  Sine  Waves — Equiva- 
lent Phase  Difference — Addition  and  Subtraction  of  Non- 
sinusoidal  Waves — 

CHAPTER  V 

CIRCUITS  CONTAINING  RESISTANCE,  INDUCTANCE  AND  CAPACITANCE.  .  115 
Coefficient  of  Self-induction  or  the  Self-inductance  of  a  Circut — 
Coefficient  of  Mutual-induction  or  the  Mutual-inductance  of  a 
Circuit — Henry,  Secohm  or  Quadrant — Energy  of  the  Field — 
Effect  of  Self-inductance  for  a  Circuit  Carrying  an  Alternating 
Current — Capacitance — Effect  of  a  Condenser  in  an  Alternating- 
current  Circuit — Circuit  Containing  Constant  Resistance  and  Con- 


CONTENTS  xi 

PAGE 

slant  Self-inductance  in  Series — Solution  of  the  Differential  Equa- 
tion of  a  Circuit  Containing  Constant  Resistance  and  Constant 
Self-inductance  in  Series — Energy  in  the  Electromagnetic  Field — 
Breaking  an  Inductive  Circuit — Impedance  and  Reactance — 
Vector  Method  of  Determining  the  Steady  Current  for  a  Circuit 
Having  Constant  Resistance  and  Constant  Self-inductance  in 
Series — Polar  Expression  for  the  Impedance  of  a  Circuit  Containing 
Constant  Resistance  and  Constant  Self-inductance  in  Series — 
Circuit  Containing  Constant  Resistance  and  Constant  Capaci- 
tance in  Series — Solution  of  the  Differential  Equation  for  a  Circuit 
Containing  Constant  Resistance  and  Constant  Capacitance  in 
Series — Energy  of  the  Electrostatic  Field — Impedance  and  React- 
ance— Vector  Method  of  Determining  the  Steady  Component  of 
the  Current  in  a  Circuit  Having  Constant  Resistance  and  Constant 
Capacitance  in  Series — Polar  Expression  for  the  Impedance  of  a 
Circuit  Containing  Constant  Resistance  and  Constant  Capacitance 
in  Series — Solution  of  the  Differential  Equation  for  a  Circuit  Con- 
taining Constant  Resistance,  Constant  Self-inductance  and  Con- 
stant Capacitance  in  Series — Vector  Method  of  Determining  the 
Steady  Component  of  the  Current  in  a  Circuit  Containing  Con- 
stant Resistance,  Constant  Self-inductance  and  Constant  Capa- 
citance in  Series — Polar  Expression  for  the  Impedance  of  a  Circuit 
Containing  Constant  Resistance,  Constant  Self-inductance  and 
Constant  Capacitance  in  Series — Equation  for  the  Velocity  and 
Displacement  of  a  Mechanical  System  Having  Friction,  Mass 
and  Elasticity. 

CHAPTER  VI 

MUTUAL- INDUCTION ^.-r.   .   .    .   173 

Mutual-induction — Coefficient  of  Mutual-induction  or  Mutual- 
inductance — Voltage  Drop  across  Circuits  Having  Resistance  and 
Self-  and  Mutual-inductance — Coefficients  of  Mutual-induction, 
Mi2  and  M2i,  of  Two  Circuits  in  a  Medium  of  Constant  Permea- 
bility are  Equal — Magnetic  Leakage  and  Leakage  Coefficients — 
Relation  among  the  Mutual-  and  Self-inductances  of  Two  Circuits 
Containing  No  Magnetic  Material,  i.e.,  Having  Constant  Magnetic 
Reluctance — Coefficient  of  Electromagnetic  Coupling  between 
Two  Electric  Circuits  Having  Mutual-inductance — General 
Equations  for  the  Voltage  Drops  Across  Two  Inductively  Coupled 
Circuits  Each  Having  Constant  Resistance,  Constant  Self-induc- 
tance and  Constant  Capacitance  in  Series — Leakage-inductance  of 
Coupled  Circuits — Relations  Among  the  Fluxes  Corresponding  to 
the  Self-inductance,  Leakage-inductance  and  Mutual-inductance — 
Voltages  Induced  in  the  Windings  of  an  Air-core  Transformer — 
Vector  Diagrams  of  an  Air-core  Transformer. 


xii  CONTENTS 

CHAPTER  VII 

PAGE 
IMPEDANCES  IN  SERIES  AND  PARALLEL,  EFFECTIVE  RESISTANCE  AND 

REACTANCE 196 

Impedances  in  Series — Complex  Method — Series  Resonance — Free 
Period  of  Oscillation  of  a  Circuit  Containing,  Resistance  Induc- 
tance and  Capacitance  in  Series — Impedances  in  Parallel — Con- 
ductance, Susceptance  and  Admittance  of  a  Circuit — Vector 
Method — Resistance  and  Reactance  in  Terms  of  Conductance  and 
Susceptance — Polar  Expression  for  Admittance — Parallel  Reson- 
ance— Impedances  in  Series-parallel — Filter  Circuits — Effect  of 
Change  of  Reactance  and  Resistance  with  Current — Non-sinus- 
soidal  Voltage  Impressed  on  a  Circuit  Containing  Constant  Imped- 
ances in  Series  and  in  Parallel — Effective  Resistance — Effective 
Reactance — Equivalent  Resistance,  Reactance  and  Impedance  of 
a  Circuit  and  Also  Equivalent  Conductance,  Susceptance  and 
Admittance  of  a  Circuit — General  Summary  of  the  Conditions 
in  Series  and  Parallel  Circuits. 

CHAPTER  VIII 

POLYPHASE  CURRENTS 245 

Generation  of  Polyphase  Currents — Double-subscript  Notation  for 
Polyphase  Circuits — Wye  and  Delta  Connections  for  Three-phase 
Generators  and  for  Three-phase  Circuits — Relative  Magnitudes 
and  Phase  Relations  of  Line  and  Phase  Currents  and  of  Line  and 
Phase  Voltages  for  a  Balanced  Three-phase  System  Having  Sinu- 
soidal Current  and  Voltage  Waves — Relative  Magnitudes  and 
Phase  Relations  of  Line  and  Phase  Currents  and  of  Line  and 
Phase  Voltages  for  a  Balanced  Four-phase  System  Having  Sinu- 
soidal Current  and  Voltage  Waves — Relative  Magnitudes  of 
Line  and  Phase  Currents  and  Line  and  Phase  Voltages  for  Balanced 
Star-  and  Mesh-connected  N-Phase  Systems  Having  Sinusoidal 
Current  and  Voltage  Waves — Balanced  and  Unbalanced  Y-con- 
nected  Loads. 

CHAPTER  IX 

KIRCHHOFF'S  LAWS  AND  EQUIVALENT  Y-  AND  A-CONNECTED  CIRCUITS.  271 
Kirchhoff's  Laws — Balanced  Y-connected  Loads  Connected  across 
Three-phase  Circuits  Having  Balanced  Voltages — Equivalent 
Unbalanced  Y-  and  A -connected  Three-phase  Circuits — Relations 
between  the  Constants  of  Unbalanced  Equivalent  A-  and  Y- 
connected  Three-phase  Circuits — Equivalent  Wye  and  Delta 
Impedances  for  Balanced  Loads — 

CHAPTER  X 

HARMONICS  IN  POLYPHASE  CIRCUITS 287 

Relative  Magnitudes  of  Line  and  Phase  Currents  and  Line  and 
Phase  Voltages  of  Balanced  Polyphase  Circuits  when  the  Currents 


CONTENTS  xiii 

PAGE 

and  Voltages  are  not  Sinusoidal — Wye  Connection — Delta  Con- 
nection— Equivalent  Wye  and  Delta  Voltages  of  Balanced  Three- 
phase  Systems  Having  Non-sinusoidal  Waves — Harmonics  in 
Balanced  Four-phase  Circuits — Harmonics  in  Balanced  Six-phase 
Circuits. 

CHAPTER  XI 

POWER    AND    POWER-FACTOR    OF    POLYPHASE    CIRCUITS,    RELATIVE 
AMOUNTS  OF  COPPER  REQUIRED  FOR  POLYPHASE  CIRCUITS,  POWER 

MEASUREMENTS  IN  POLYPHASE  CIRCUITS 314 

Power  and  Power-factor  of  Balanced  Polyphase  Circuits — Power- 
factor  of  an  Unbalanced  Polyphase  Circuit — Relative  Amounts  of 
Copper  Required  to  Transmit  a  Given  Amount  of  Power  a  Fixed 
Distance,  with  a  Fixed  Line  Loss  and  Fixed  "Voltage  between  Con- 
ductors, over  a  Three-phase  Transmission  Line  under  Balanced 
Conditions  and  over  a  Single-phase  Line — Relative  Amounts  of 
Copper  Required  to  Transmit  a  Given  Amount  of  Power  a  Fixed 
Distance,  with  a  Fixed  Line  Loss  and  a  Fixed  Voltage  between 
Conductors,  over  a  Four-phase  Transmission  Line  under  Balanced 
Conditions  and  over  a  Single-phase  Line — Relative  Amounts  of 
Copper  Required  to  Transmit  a  Given  Amount  of  Power  a  Fixed 
Distance,  with  a  Fixed  Line  Loss  and  a  Fixed  Voltage  to  Neutral, 
when  the  Loads  are  Balanced — Power  Measurements  in  Three- 
phase  Circuits — Proof  of  the  Three-wattmeter  Method  for  Measur- 
ing the  Power  in  a  Three-phase  Circuit — The  N-wattmeter  Method 
for  Measuring  the  Power  in  an  N-phase  Circuit — Two-wattmeter 
Method  for  Measuring  Power  in  a  Three-phase  Circuit  and  the 
(N-l)-wattmeter  Method  for  Measuring  Power  in  an  N-phase 
Circuit — Relative  Readings  on  Balanced  Loads  of  Wattmeters 
Connected  for  the  Two-wattmeter  Method  of  Measuring  Power  in 
a  Three-phase  Circuit  when  the  Current  and  Voltage  Waves  are 
Sinusoidal — Determination  of  the  Power-factor  of  a  Balanced 
Three-phase  Circuit,  when  the  Current  and  Voltage  Waves  are 
Sinusoidal,  from  the  Readings  of  Two  Wattmeters  Connected 
to  Measure  the  Total  Power  by  the  Two-wattmeter  Method — 
Measurement  of  the  Reactive  Power  of  a  Balanced  Three-phase 
Circuit  whose  Current  and  Voltage  Waves  are  Sinusoidal. 

CHAPTER  XII 

UNBALANCED  THREE-PHASE  CIRCUITS. 337 

Unbalanced  Circuits — Direct,  Reverse  and  Uniphase  Components 
of  Three-phase  Vectors — Determination  of  the  Direct,  Reverse 
and  Uniphase  Components  of  a  Three-phase  System — A  Simple 
Graphical  Construction  for  Finding  the  Direct-  and  Reverse-phase 
Components  of  a  Three-phase  Circuit  whose  Vectors  are  Sinusoidal 


xiv  CONTENTS 

PAGE 

and  Contain  no  Uniphase  Components — Mutual-induction 
between  a  Three-phase  Transmission  Line  and  a  Neighboring  Tele- 
phone Line — Power  in  an  Unbalanced  Three-phase  Circuit  when 
the  Currents  and  Voltages  are  Sinusoidal — Phase  Balancer — 
Copper  Loss  in  an  Unbalanced  Three-phase  Circuit  in  Terms  of  the 
Direct,  Reverse  and  Uniphase  Components  of  the  Currents — 
Effect  of  Impressing  an  Unbalanced  Voltage  on  a  Three-phase 
Alternating-current  Motor  or  Generator. 

CHAPTER  XIII 

REACTANCE  OF  A  TRANSMISSION  LINE 355 

Reactance  of  a  Single-Phase  Transmission  Line — Average  Reac- 
tance per  Conductor  of  a  Completely  Transposed,  Ungrounded, 
Three-phase  Transmission  Line — Transfer  of  Power  among  the 
Conductors  of  a  Three-phase  Transmission  Line — Mutual-induc- 
tion between  Transmission  Lines  or  between  a  Transmission  Line 
and  a  Telephone  Line — Voltage  Induced  in  a  Telephone  Circuit  by 
a  Three-phase  Transmission  Line  which  is  Parallel  to  the  Telephone 
Line — Voltage  Induced  in  a  Telephone  Line  by  an  Adjacent  Three- 
phase  Transmission  Line  which  Carries  an  Unbalanced  Load — 
Third-harmonic  Voltage  Induced  in  a  Telephone  Line  which  is 
Adjacent  to  a  Transmission  Line  that  is  Fed  from  a  Bank  of 
Transformers  which  are  Connected  in  Wye  on  Both  Primary  and 
Secondary  Sides. 

CHAPTER  XIV 

CAPACITANCE  OF  A  TRANSMISSION  LINE 376 

Capacitance  of  Two  Straight  Parallel  Conductors — Charging 
Current  of  a  Transmission  Line — Capacitance  of  a  Balanced 
Three-phase  Transmission  Line  with  Conductors  r,t  the  Corners 
of  an  Equilateral  Triangle,  Neglecting  the  Effect  of  the  Earth — 
Voltage  Induced  in  a  Telephone  or  Telegraph  Line  by  the  Electro- 
static Induction  of  the  Charges  on  the  Conductors  of  an  Adjacent 
Transmission  Line. 

CHAPTER  XV 

SERIES-PARALLEL    CIRCUITS    CONTAINING    UNIFORMLY    DISTRIBUTED 

RESISTANCE,  REACTANCE,  CONDUCTANCE  AND  SUSCEPTANCE  .  .  .  405 
Series-parallel  Circuits — Equations  for  Voltage  and  Current  of  a 
Line  whose  Series  Resistance  and  Reactance  per  Unit  Length  of 
Line  and  whose  Parallel  Conductance  and  Susceptance  per  Unit 
Length  of  Line  are  Constant — Propagation  Constant,  Attenuation 
Constant,  Wave-length  Constant,  Velocity  of  Propagation  and 
Length  of  Line  in  Terms  of  Wave  Length — Evaluation  of  the  Equa- 
tions for  Current  and  Voltage  when  Tables  of  Hyperbolic  Functions 
of  Complex  Quantities  are  not  Available. 

INDEX.  .   425 


J     J      \ 

'J  »   "» 


PRINCIPLES  OF 
ALTERNATING  CURRENTS 


CHAPTER  I 

ALGEBRA    OF   VECTORS   AND  OF   COMPLEX   QUANTITIES    USED 
IN  ELECTRICAL  ENGINEERING 

Quantities  Involved  in  the  Solution  of  Problems  in  Alternating 
Currents. — All  quantities  involved  in  the  solution  of  ordinary 
problems  in  direct  currents  are  simple  algebraic  quantities. 
All  equations  are  simple  algebraic  equations  and  may  be  handled 
by  any  of  the  ordinary  algebraic  methods..  These  statements  are 
equally  true  when  applied  to  the  currents,  voltages,  power,  etc., 
-  existing  at  any  instant  of  time  in  an  alternating-current  circuit, 
i.e.,  when  applied  to  the  so-called  instantaneous  values  of  current, 
voltage  and  power.  Except  in  special  cases,  instantaneous 
values  are  not  important.  What  is  desired  is  the  average  power 
and  the  effective  voltage  and  the  effective  current.  Effective 
voltage  and  effective  current  are  not  simple  algebraic  quantities 
and  cannot  be  handled  by  ordinary  algebraic  methods.  They 
are  vectors  and  must  be  handled  by  methods  which  are  applicable 
to  vectors.  For  this  reason  a  knowledge  of  vector  algebra  and 
the  algebra  of  complex  quantities  is  necessary  to  an  electrical 
engineer. 

Types  of  Vectors  met  in  Electrical  Engineering  Problems. — 
There  are  two  types  of  vectors,  both  of  which  occur  in  many 
alternating-current  problems.  There  are  vectors  which  are 
fixed  in  direction  in  space,  i.e.,  space  vectors,  and  vectors  which 
are  constant  in  magnitude  and  revolve  with  a  constant  angular 
velocity,  i.e.,  revolving  vectors.  The  latter  are  sometimes  called 
time  vectors.  A  constant  force  is  a  good  example  of  a  space 
vector.  It  acts  in  a  fixed  direction  in  space  with  a  constant 
intensity  and  may  be  represented  both  in  direction  and  magnitude 

1 


ALTERNATING  CURRENTS 


l>v  $  $t?aigktJi?iQ  wjio&e  tefcgth  and  direction  represent,  respective- 
ly} iKe  ^ma^ttwJe^amT  the"  direction  of  the  force.  An  alternating 
current,  which  varies  sinusoidally  with  time,  may  be  represented 
by  a  revolving  vector  of  fixed  magnitude  which  revolves  with  a 
constant  angular  velocity.  If  the  length  of  the  vector  represents 
the  maximum  value  of  the  current,  its  projection  on  a  fixed 
reference  axis  is  the  value  of  the  current  at  the  instant  of  time 
considered.  The  number  of  revolutions  per  second  made  by  the 
revolving  line  is  equal  to  the  number  of  cycles  gone  through  by 
the  current  per  second. 

The  phase  difference  between  two  vector  quantities  is  the 
angle  between  the  two  vectors  which  represent  the  quantities. 

Revolving  vectors  may  be  handled  by  any  of  the  processes 
which  are  applicable  to  space  vectors  by  merely  considering 
them  at  some  particular  instant  of  time  or  they  may  be  treated 
by  methods  which  are  applicable  to  them  alone. 

Solution  of  Problems  Involving  Vectors  by  the  Methods  of 
Trigonometry. — Vectors  may  be  added  or  subtracted  by  the 
use  of  trigonometrical  formulas  for  the  solution  of  triangles  but, 
when  there  are  more  than  two  vectors,  this  method  of  addition 
or  subtraction  becomes  unnecessarily  long  and  cumbersome  and 
when  applied  to  any  but  the  simplest  problems  becomes  hope- 
lessly involved. 

Vector  Algebra. — The  vector  algebra  which  is  necessary  for 
handling  problems  in  alternating  currents  is  comparatively 
simple.  It  makes  easy  the  solution  of  alternating-current 
problems  which  would  otherwise  be  difficult.  A  knowledge  of 
vector  algebra  is  therefore  one  of  the  most  useful  tools  to  the 
electrical  engineer. 

There  are  two  ways  of  treating  vectors.  They  may  be  referred 
to  rectangular  coordinate  axes  and  expressed  in  terms  of  their 
components  along  these  axes  or  they  may  be  expressed  in  terms 
of  polar  coordinates.  Each  of  these  methods  has  its  advantages 
and  both  are  useful.  The  former  is  better  for  addition  and 
subtraction  of  vectors,  but  can  be  used  for  multiplication  and 
division  of  vectors  as  well.  The  latter  is  better  when  only  multi- 
plication and  division  are  to  be  performed.  It  cannot  be  used 
for  addition  and  subtraction.  To  change  the  expression  of  a 
vector  from  one  form  to  the  other  is  a  simple  matter. 


VECTORS  AND  COMPLEX  QUANTITIES  3 

Method  of  Complex  Quantities. — The  method  of  handling 
vectors  when  they  are  referred  to  coordinate  axes  is  known  as 
the  Method  of  Complex  Quantities.  In  this  method  all  vectors 
are  resolved  into  two  components,  respectively  along  and  at 
right  angles  to  some  conveniently  chosen  axis  of  reference.  An 
operator  j  is  attached  to  the  component  at  right  angles  to  the 
axis  of  reference  to  distinguish  it  from  the  component  along  that 
axis.  The  name  "Complex"  as  applied  to  this  method  does  not 
indicate  complexity  of  method.  The  method,  so  far  as  its 
application  is  concerned,  should  be  called  the  "  Simplex  Method. " 
The  name  " complex"  comes  from  the  fact  that  each  vector 
involved  is  resolved  into  a  so-called  real  and  imaginary  compo- 
nent, neither  of  which,  however,  is  actually  imaginary. 

The  components  to  which  j  is  attached  are  called  the  imaginary 
components  or  simply  "  imaginaries. "  The  other  components 
are  called  the  real  components  or  simply  "reals."  The  two 
rectangular  axes  along  which  these  components  lie  are  called  the 
axis  of  imaginaries  and  the  axis  of  reals.  The  axis  of  reals  is  the 
axis  from  which  the  angles  are  measured 
which  show  the  phase  relations  of  the 
vectors.  Although  j  =  \/  —  1  is  an 
imaginary  quantity,  neither  the  com- 
ponent of  the  vector  to  which  j  is 
attached  nor  the  axis  along  which  it 
lies  is  imaginary.  Both  are  real,  just 
as  real  as  the  other  component  and  the 
other  axis. 

Let  OA,  Fig.  1,  be  a  vector  A  making  any  angle  0  with  the 
reference  axis  OX. 

Let  the  vector  OA  =  A  be  resolved  into  two  components,  OB.  = 
B  and  OC  =  C  respectively  along  and  at  right  angles  to  the 
axis  OX.  The  vector  has  for  its  magnitude  \/B2  +  C2  and  makes 

£• 

an  angle  tan"1  ^  with  OX.     The  expression  for  the  vector  A, 
in  terms  of  its  components,  may  be  written 

A  =  B  +  C  (1) 

where  the  addition  must  be  considered  in  a  vector  sense.     To 
make  it  possible  to  distinguish  between  vector  and  non-vector 


4  PRINCIPLES  OF  ALTERNATING  CURRENTS 

quantities  in  equations,  it  is  customary  to  use  a  short  line  or  a  dot 
over  letters  or  numbers  representing  vector  quantities.  For 
example 

A  =  B  +  C  (2) 

A  =  B  +  C  (3) 

Sometimes  the  dot  is  placed  under  the  vector  instead  of  over  it. 

Since  nearly  all  expressions  met  in  alternating  current  problems 
are  vector  expressions,  the  use  of  dots  or  dashes  is  often  unneces- 
sary. They  are  frequently  omitted.  The  simple  algebraic 
expressions  which  occur  are  easily  distinguishable  from  the 
vector  expressions  without  the  use  of  any  special  symbols. 

Operator  j. — Some  notation  must  be  adopted  which  will  make 
it  possible  to  distinguish  readily  between  the  components  along 
the  two  axes.  The  letter  j  is  used  for  this  purpose.  The  letter  j 
is  an  operator  which  indicates  that  a  vector  to  which  it  is  attached 
has  been  rotated  through  ninety  degrees  in  a  positive  direction. 
Counter-clockwise  direction  is  always  considered  positive  and  the 
horizontal  direction  is  usually  taken  for  the  axis  of  reference. 
Left  to  right  along  this  axis  is  considered  positive 

The  operator  j  does  not  differ,  except  in  the  effect  it  produces, 
from  other  common  operators,  such  as  plus  and  minus  signs, 
multiplication  and  division  signs,  exponents  and  radical 
signs,  log,  sin,  cos,  etc.  For  example,  the  exponent  3  with  A3  is 
merely  an  abbreviated  way  of  writing  A  X  A  X  A .  The  exponent 
3  is  an  operator  that  indicates  that  a  certain  operation  is  to  be  per- 
formed on  A,  namely,  that  it  is  to  be  multiplied  by  itself  twice. 
In  a  similar  way  the  operator  j  indicates  that  the  vector  to  which 
it  is  attached  has  been  rotated  through  ninety  degrees  in  a 
counter-clockwise  direction.  Using  the  operator  j,  equation  (2) 
becomes 

A  =  B  +  JC  (4) 

B  +  jC  is  one  form  of  vector  expression  for  the  vector  A . 
C,  or  in  general  the  part  of  a  vector  to  which  j  is  attached,  is  the 
component  at  right  angles  to  the  reference  axis,  and  B,  or  the 
component  without  j,  is  the  component  along  the  reference  axis. 
C  without  the  j  attached  would  lie  in  a  positive  direction  along 
the  reference  axis.  The  letter  j  indicates  that  it  has  been  rotated 
through  ninety  degrees  in  a  positive  direction  from  the  axis  of 


VECTORS  AND  COMPLEX  QUANTITIES  5 

reference.  Since  the  OX  or  horizontal  axis  was  taken  as  the 
axis  of  reference,  jC  lies  vertically  Upward  or  along  the  OF  axis. 
The  expression  given  for  the  vector  A  in  equation  (4)  is  known  as 
its  complex  expression.  The  reason  for  the  term  complex  will  be 
explained  later.  It  does  not  indicate  complexity  of  expression. 
It  has  other  significance. 

If  the  operator  j  rotates  a  vector  to  which  it  is  attached 
through  +  90  degrees,  applying  j  twice  or  f  once  should  rotate  it 
through  +  180  degrees.  Applying  j  twice  or  j2  once  reverses  a 
vector  and  is  equivalent  to  multiplying  the  vector  by  —  1. 
Therefore 

j  X  j  =  f  =  -  1 
and 


Applying  j  three  times  or  j3  once  gives 

j*  =j  Xj2  =JX  (-1)  =  -j 

The  operator*  —j  therefore  rotates  any  vector  to  which  it  is 
attached  through  —90  degrees  or  through  90  degrees  in  a  clock- 
wise or  negative  direction. 
Representation  of  Vectors  by  the  Use  of  the  Operator  j. — The 


four  vectors  A,  jA,  j2A  =  —A 
Fig.  2. 


-A 


and  j3A  =  —jA  are  shown  in 


FIG.  2. 

Four  vectors  of  equal  magnitude  are  given  by  the  following 
equations. 

Ii  =  a  +  jb  (5) 

A*  =  -a+jb  (6) 

A*  =  -a  -  jb  (7) 

I4  =  a  -  jb  (8) 


6 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


These  have  magnitudes  of  A  =  \/a2  +  b2  and  lie  respectively  in 
the  first,  second,  third  and  fourth  quadrants.  They  are  shown 
in  Fig.  3.  Equations  (5),  (6),  (7)  and  (8)  are  the  complex  ex- 
pressions for  the  vectors. 


FIG.  3. 


The  four  vectors  make  angles,  with  the  axis  of  reference,  of 


«2, 


tan  a\  =  - 


and 

6 

a 


where 


sin  «i  = 


tan  «2  = 


tan  0:3  = 


tan  «4  = 


+ 
b 


sin  «2  =  — 7= 


—  a 
-b 
—a 
-b 


sm 


sm  a4  = 


-•& 


b2 


-b 


COS  tt2  = 


COS  <*3    = 


COS  «4    = 


—  7=  (11) 

Va2  +  b2 


+~b 


(12) 


When  expressing  the  value  of  the  tangent  of  the  angle  «, 
which  a  vector  makes  with  the  reference  axis  OX,  it  should  be 
left  in  the  form  of  a  ratio  with  the  proper  signs  attached  to  both 
numerator  and  denominator.  Unless  this  is  done  it  is  impossible 
to  tell  in  which  of  two  quadrants  the  angle  lies.  Neither  the  sine 
nor  the  cosine  of  the  phase  angle  is  alone  sufficient  to  fix  the 
position  of  a  vector.  For  example:  referring  to  equations  (9), 

(10),  (11)  and  (12),  sin  a  =      ,  —       =  might  refer  to  an  angle  in 

V  a2  +  b2 

either  the  third  or  the  fourth  quadrant  and  cos  a  = 


^= 
a2  +  b2 

might  indicate  an  angle  in  either  the  second  or  the  third  quadrant. 

Real  and  Imaginary  Components  of  a  Vector  and  Real  and 

Imaginary  Axes.  —  The  two  parts  a  and  b  of  the  vector  expression 


VECTORS  AND  COMPLEX  QUANTITIES  7 

A  =  a  +  ft  are  called  respectively  the  real  part  and  the  im- 
aginary part  of  the  vector.  The  two  axes  along  which  they  lie  are 
called  the  axis  of  reals  and  the  axis  of  imaginai  ies.  The  expres- 
sion a  +  jb  is  called  the  complex  expression  of  the  vector  A. 
Neither  the  imaginary  part,  6,  of  the  vector  nor  the  axis  of 
imaginaries,  OF,  is  imaginary.  Both  are  just  as  real  as  the  com- 
ponent a,  and  just  as  real  as  the  axis  of  reals.  The  term  im- 
aginary comes  from  the  mathematical  significance  of  the  operator 
j  which  is  attached  to  the  so-called  imaginary  component. 
Mathematically,  the  operator  j  =  \/  —  1  is  an  imaginary  quan- 
tity. When  used  as  a  rotating  operator  it  does  not  make  the 
component  of  a  vector  to  which  it  is  attached  any  less  real  than 
the  so-called  real  component. 

Operator  (cos  a  ±  j  sin  a). — The  four  vectors  AI,  Az,  A3  and 
A  4,  given  in  equations  (5),  (6),  (7)  and  (8),  page  5  and  shown  in 
Fig.  3,  may  be  written 

Ai  =  Ai(cos  «i  +  j  sin  «i)  (13) 

A.2  =  A 2 (cos  «2  +  j  sin  «2)  (14) 

Az  =  A3(cos  «3  +  j  sin  03)  (15) 

A4  =  A4(cos  ou  +  j  sin  «4)  (16) 

In  all  the  equations  the  angles  a  are  positive  angles,  i.e., 
they  are  measured  in  a  positive  direction  from  the  axis  of 
reference. 

If  a,  =  0, 

A i  =  AKcos  0  +  j  sin  0)  =  A^l  +  JO) 

A  i  would  lie  therefore  in  a  positive  direction  along  the  axis  of 
reals.  When  «i  has  any  value  other  than  zero,  such  as  0,  the 
vector  would  make  a  positive  angle  0  with  the  axis  of  reals, 
(cos  0  +  j  sin  0)  is  therefore  an  operator  which  will  rotate  the 
vector  A  i  through  a  positive  angle  0.  In  general 

(cos  a  -f-  j  sin  a)  (17) 

is  an  operator  which  rotates  a  vector  to  which  it  is  applied  through 
a  positive  angle  a.  It  makes  no  difference  whether  the  vector 
to  which  the  operator  is  applied  lies  along  the  axis  of  reals  or  in 
any  other  direction,  the  operator  will  rotate  it  from  its  original 
position  through  a  degrees  in  a  positive  or  counter-clockwise 
direction. 


8  PRINCIPLES  OF  ALTERNATING  CURRENTS 

A  little  consideration  will  show  that 

(cos  a  —  j  sin  a)  (18) 

is  an  operator  which  rotates  a  vector  to  which  it  is  applied  through 
an  angle  a  in  a  negative  or  clockwise  direction  from  its  original 
position. 

In  both  of  the  operators  (cos  a  +  j  sin  a)  and  (cos  a  —  j  sin  a) 
the  numerical  value  of  cosine  and  sine  must  be  taken  for  the 
positive  angle  a.  Whether  positive  or  negative  rotation  is 
produced  depends  on  the  sign  attached  to  j  in  the  operator. 
Although  the  sine  and  cosine  are  for  the  positive  angle  a,  it  must 
not  be  forgotten  that  the  sine  and  cosine  of  certain  angles  are 
negative. 

The  general  operator  is  (cos  a  ±  j  sin  a). 

If  a  =0,  cos  a  =  1  and  sin  a  =  0, 

and  (cos  a.  ±  j  sin  a)  =  1. 
If  a  =    90°,  cos  a  =  0  and  sin  a  =  1, 

and  (cos  a  ±  j  sin  a)  =  ±j. 
If  a  =  180°,  cos  a  .=  —  1  and  sin  a  =  0, 

and  (cos  a  ±  j  sin  a)  =  —  1. 
If  a  =  270°,  cos   a  =  0  and  sin  a  =  -1, 

and  (cos  a  ±  j  sin  a)  =  +j. 

Operator  which  Rotates  the  Reference  Axes  through  an  Angle 

a. — In  certain  cases  it  is  necessary  to  refer  a  vector  to  a  new 
axis  of  reference  which  is  displaced  from  the  original  axis  by  some 
definite  angle,  such  as  a.  Rotating  the  axes  with  respect  to 
a  vector  is  equivalent  to  rotating  the  vector  in  the  opposite 
direction  with  respect  to  the  axes.  Therefore,  if  the  operator 
(cos  a  +  j  sin  a)  rotates  a  vector  through  a  positive  angle  a 
with  respect  to  the  axes,  the  same  operator  may  be  considered 
to  rotate  the  axes  through  a  negative  angle  a  with  respect  to 
the  vector. 

The  operators  (cos  a  —  j  sin  a)  and  (cos  a  +  j  sin  a)  are 
therefore  two  operators  which  applied  to  a  vector  will  rotate 
the  axes  with  respect  to  it  through  an  angle  a  respectively  in  a 
positive  and  in  a  negative  direction.  In  problems  there  will  fre- 
quently occur  vectors  which  are  referred  to  different  reference 
axes.  Before  these  vectors  can  be  added  or  subtracted,  multi- 
plied or  divided  they  must  all  be  referred  to  the  same  reference 


VECTORS  AND  COMPLEX  QUANTITIES  9 

axes.  This  can  be  easily  done,  provided  the  angle  between  their 
reference  axes  is  known,  by  applying  the  operator  (cos  a  +  j  sin  a). 
The  operator  with  the  negative  sign  produces  a  positive  or  counter 
clockwise  rotation  of  the  axes.  With  the  positive  sign  the  opera- 
tor produces  a  negative  or  clockwise  rotation  of  the  axes. 

Successive  Application  of  Rotating  Operators,  Powers  and 
Roots  of  Operators,  Reciprocal  of  an  Operator.  —  Consider  the 
two  operators 

ki  =  cos  0i  +  j  sin  0\ 

k2  =  cos  02  -h  j  sin  02 

They  rotate  a  vector  to  which  they  are  applied  through  angles  0i 
and  02  respectively.  Applied  in  succession  they  should  rotate 
it  successively  through  angles  0i  and  02  or  through  a  total  angle 
(0i  +  02). 

fci  X  k2  =  (cos  0i  +  j  sin  0i)(cos  02  +  j  sin  02) 
=  (cos  0i  cos  02  —  sin  0i  sin  02) 

-h  ./(sin  Si  cos  02  +  cos  0i  sin  02) 
=  cos  (0i  +  02)  +  j  sin  (0!  +  02)  (19) 

The  product  of  two  operators  is  thus  a  new  operator  which 
produces  a  rotation  equal  to  the  algebraic  sum  of  the  rotations 
produced  by  the  operators  individually.  Similarly  the  product 
of  any  number  of  rotating  operators  is  a  new  operator  which 
produces  a  rotation  equal  to  the  algebraic  sum  of  the  rotations 
produced  by  the  operators  separately. 
kiXk2Xk3X  .  .  X  kn  =  (cos  0i  +  j  sin  00 

X  (cos  02  +  j  sin  02) 
X  (cos  03  +  j  sin  03) 
X     .    .    X    (cos  0n  +  j  sin  0n) 

=    COS    (01    +    0*   +    03   +    •       •      +   0n) 

+  j  Sin    (01   +    02   +    03   +    •      •      +    On) 

=  cos  20  +  jsin  20  (20) 

If  0i,  02,  03,  etc.  are  all  equal 

ki  X  k2  X  k3  X    .    .    .    X  kn  =  kn  =  cos  (nO)  +  j  sin  (n8) 
If  n0  =    180  degrees  or  IT  radians 


kn  =  k9  =  COSTT  +j  shiTT  =  —1  (21) 

2 

*  =  \/  -  1  =  /  =  cos     +  j  sin  (22) 


10  PRINCIPLES  OF  ALTERNATING  CURRENTS 

2 

Therefore  \/  —  I  =  j  n  is  an  operator  which  rotates  a  vector 
through  -  radians.  Since  the  operator  •%/—  1  is  equal  to  the 

nth  root  of  minus  one  it  must  have  n  roots  or  values,  each  of 
which  will  produce  a  definite  rotation.     One  of  these  roots  is 

(cos  —  h  j  sin  -J  ,  which  produces  the  rotation  of     radians. 

Adding  any  number  of  ±  2ir  radians  to  an  angle  does  not  alter 
the  value  of  its  sine  or  of  its  cosine.  Equation  (21)  may  therefore 
be  written 

—  1  =  cos  TT  +  j  sin  TT 

=  cos  (2q  +  !)TT  +  j  sin  (2q  +  I)TT 

where  q  is  any  positive  or  negative  integer. 


f  =  i  =  cos  V  +  jsin  r  (23) 

\  9V  I  \  Tt  / 

2 

It  might  appear  from  equation  (23)  that  the  operator  jn  has 
an  infinite  number  of  values,  since  q  may  have  an  infinite  number 
of  values.  There  are,  however,  only  n  different  roots  of  minus 
one.  After  the  nth  root,  the  roots  repeat. 

For  example,  let  n  =  3.  In  this  case  there  are  only  three 
different  roots.  These  are: 

For  q  =  0,     j%  =  -   radians  or  60  degrees. 

o 

For  q  =  1,     j**  =  TT   radians  or  180  degrees. 
For  q  =  2,     j**  =  -  TT  radians  or  300  degrees. 

o 

For  any  greater  values  of  q  the  roots  repeat.     For  example, 

for  q  =  3,  the  root  is  ~  TT,  which  is  equivalent  to  ~.     This  is  the 
o  o 

same  as  the  first  root.     If  q  is  given  negative  values,  the  same 
three  roots  will  result.     For  example,  if  q  =  —1,  the  root  is  —  « 

radians  or  —  60  degrees.     This  is  the  same  as  plus  300  degrees  or 

is  the  same  as  the  third  root  obtained  using  positive  values  of  q. 

The    operator  j  =  -\/  —  l   has  two  roots,  which  produce  re- 

spectively   -f-90    and    —90    degrees  rotation,     When  using  j, 


VECTORS  AND  COMPLEX  QUANTITIES  11 

however,  as  an  operator  in  the  complex  expression  for  a  vector, 
the  positive  root  is  arbitrarily  used.  In  such  expressions  j  is 
used  as  an  operator  which  produces  a  rotation  of  plus  ninety 
degrees.  Although  the  operator  j  is  universally  employed  in  work 
involving  alternating  currents  to  produce  a  rotation  of  +90 
degrees,  the  roots  of  j  which  produce  rotations  of  fractional 
parts  of  90  degrees  are  not  employed,  as  other  more  convenient 
forms  of  operator,  which  are  single  valued,  are  available  for  this 
purpose. 

Reciprocal  of  the  Operator  (cos  a  ±  j  sin  a). — Consider  the 
reciprocal  of  the  operator  (cos  a  +  j  sin  a),  which  produces  a 
positive  rotation  of  a  degrees. 

1 1 cos  a  —  j  sin  a 

cos  a  +  j  sin  a       cos  a  +  j  sin  a       cos  a  —  j  sin  a 

cos  a  —  j  sin  a  .   .  /CM, 

=  — ^ —         .   , —  =  cos  a  —  j  sin  a      (24) 
cos2  a  +  sin2  a 

Therefore,  the  reciprocal  of  an  .operator  which  produces  a 
rotation  of  a  is  an  operator  which  produces  a  rotation  of  —a. 
Dividing  a  vector  by  an  operator  which  produces  a  rotation  of 
any  angle  a  gives  the  same  result  as  multiplying  it  by  an  operator 
which  produces  a  rotation  of  a  in  the  opposite  direction. 

In  general 

(cos  ai  +  j  sin  qQ  X  (cos  a2  +  j  sin  a2)  X 

(cos  |8i  +  j  sin  ft)  X  (cos  02  +  j  sin  02)  X 

X  (cos  an  -f  j  sin  «»)  _ 
X  (cos  /3n+j  sin  ftO  " 
cos  (or0  —  0o)  +  j  sin  (or0  —  0o) 

where  «0  is  the  algebraic  sum  of  the  angles  a,  and  00  is  the 
algebraic  sum  of  the  angles  |8. 

Operators  which  Produce  Uniform  Angular  Rotation. — Let  A 
be  any  vector  of  constant  magnitude  which  rotates  at  a  uniform 
angular  velocity  of  2wf  =  co  radians  per  second.  The  number 
of  revolutions  made  by  the  vector  per  second  is/.  Let  time,  I,  be 
reckoned  in  seconds  and  let  it  be  considered  zero  when  the 
vector  A  lies  along  the  axis  of  reference  in  a  positive  direction. 
At  any  instant  of  time  t  seconds  after  t  =  0,  the  vector  will  have 
rotated  through  ut  radians  and  will  make  an  angle  of  wt  radians 


12  PRINCIPLES  OF  ALTERNATING  CURRENTS 

with  the  axis  of  reference,  i.e.,  the  axis  of  reals.  The  operator 
which  will  produce  this  rotation  is  (cos  cot  +  j  sin  co£).  By  making 
use  of  this  operator  the  vector  A  may  be  represented  by 

A  =  A  (cos  co*  +  j  sin  coZ)  =  Aj*°l  =  Aj*ft~ 

The  operator  (cos  cot  +  j  sin  wt)  produces  a  uniform  rotation 
of  2irf  =  co  radians  per  second. 

Solution  of  Vector  Equations  when  the  Vectors  and  Complex 
Quantities  Involved  are  Expressed  in  the  Complex  Form, 
i.e.,  in  the  Form  a  +  jb. — In  any  vector  equation  the  algebraic 
sum  of  the  real  terms  on  one  side  of  the  equation  must  be  equal 
to  the  algebraic  sum  of  the  real  terms  on  the  other  side  of  the 
equation,  and  similarly  the  algebraic  sum  of  the  imaginary  terms 
on  one  side  of  the  equation  must  be  equal  to  the  algebraic  sum 
of  the  imaginary  terms  on  the  other  side  of  the  equation.  Since 
the  magnitude  of  a  vector  is  equal  to  the  square  root  of  the  sum 
of  the  squares  of  the  real  and  the  imaginary  parts,  the  square 
of  the  sum  of  the  real  terms  plus  the  square  of  the  sum  of  the 
imaginary  terms  on  one  side  of  a  vector  equation  must  be  equal 
to  the  square  of  the  sum  of  the  real  terms  plus  the  square  of  the 
sum  of  the  imaginary  terms  on  the  other  side  of  the  equation. 
The  operator  j  is  omitted  in  taking  the  squares  as  it  is  not  a  part 
of  the  magnitude  of  any  component  to  which  it  is  attached. 

In  a  direct-current  circuit,  current  is  given  by  voltage  divided 
by  resistance.  A  similar  expression  holds  for  the  current  in  an 
alternating-current  circuit.  In  the  alternating-current  circuit 
the  resistance  must  be  replaced  by  the  impedance.  If  the  circuit 
is  inductive  the  numerical  value  of  the  impedance  is  \/r2  +  x2, 
where  r  is  the  resistance  of  the  circuit  and  x  is  the  reactance  and  is 
equal  to  the  inductance  of  the  circuit  multiplied  by  2w  times  the 
frequency  of  the  impressed  voltage. 

The  current  multiplied  by  the  resistance  is  the  voltage  drop 
in  the  circuit  caused  by  the  resistance.  It  is  a  vector  which  is 
in  phase  with  the  current.  The  current  multiplied  by  the  react- 
ance is  a  voltage  drop  in  the  circuit  caused  by  the  reactance. 
It  is  a  vector  which  is  in  quadrature  with  the  current  and  lead- 
ing it.  Impedance  is  a  complex  quantity  and  its  complex 
expression  is  r  +  jx. 

V  =  l(r  +  jx) 


VECTORS  AND  COMPLEX  QUANTITIES 


13 


from  which  we  have 


7  = 


r  +  jx 

Let  V  =  100  +  j50  and  let  the  impedance  be  4  +  j3.     Then 
the  magnitude  of  I  is  • 

V(TOO)2  +  (50) 
7  =—  -== 


111.8 


V(4)2  +  (3)2 


=  22.36  amperes. 


The  vector  expression  for  the  current  is 
j      100  +  J50 
4  +J3 

This  must  be  rationalized,  to  get  rid  of  the  j  in  the  denominator, 
by  multiplying  both  numerator  and  denominator  by  the  denomi- 
nator with  the  sign  of  its  j  term  reversed. 

j       100  +  J50       4  -  J3 
4  +  J3        -  4  -  J3 
=  400  -  J30Q  +  J200  +  150 

16  -  J12  +~J12  +  9 
=  550         100 

25       J  25  3 

The  component  of  the  current  J  which  is  along  the  axis  to  which 
V  is  referred  is  22  and  the  component  at  right  angles  to  the  axis 
is  4.  The  magnitude  of  7  is  V(22)2  +  (4)2  =  22.36  amperes. 

—  4 
The  vector  7  makes  an  angle  tan  ~x  -^o" 

50 


the  reference  axis. 


0V  =  tan~] 


100' 


ev  =  +26.6  degrees. 


The  vector  diagram  for  7  and  V  is  shown  in  Fig.  4. 


=  22-j4 


FIG.  4. 


0/     =  tan- 


-4 
"22 ' 


=   - 10.3  degrees. 


=  ev  -  e,  =  26.6  -  (-10.3)  =  36.9  degrees. 


14  PRINCIPLES  OF  ALTERNATING  CURRENTS 

It  frequently  happens,  in  solving  a  vector  equation,  that  the 
magnitude  and  not  the  phase  angle  of  one  of  the  terms  is  known. 
In  such  a  case  the  equation  cannot  be  solved  as  a  vector  equation. 
If,  however,  the  term  which  cannot  be  expressed  in  vector  form 
can  be  separated  from  the  others,  the  equation  may  be  solved 
by  turning  it  into  a  simple  algebraic  equation  by  equating  the 
square  of  the  sum  of  the  real  terms  plus  the  square  of  the  sum 
of  the  imaginary  terms  on  one  side  of  the  equation  to  the  square 
of  the  term  which  cannot  be  expressed  in  vector  form,  which  is 
on  the  other  side.  Consider  the  following  example  in  which  V  is 
taken  as  the  axis  of  reference,  i.e.,  as  the  axis  of  reals. 
6000(cos  0  +  j  sin  9)  =  F(cosT)0  +  j  sin  0°)  + 

50(cos  30°  +  j  sin  30°)  (3  +  j4)     (26) 

The  angle  6  is  unknown.  Substituting  the  values  for  the 
sines  and  cosines  of  the  known  angles  in  equation  (26)  gives 
6000(cos  S+j  sin  6)  =  V(l  +  jO)  + 

50(0.866  +  j0.500)(3  +  j4) 
=  (V  +  29.9)  +  j(248.2  +  0)  (27) 

(6000)2  =  (V  +  29.9)2  +  (248.2)2 
V  +  29.9  =  -v/35,938,397  =  ±5995 

V  =  +  5965  or  -6025 

Putting  the  positive  value  of  V  in  equation  (27)  gives 
5965  +  29.9    ,    .  248.2 


=  0.9992  +  jO.0414 
cos  0  =  0.9992  and  sin  0  =  0.0414 

0  =  +2.37  degrees. 
Using  the  negative  value  of  V  gives 

cos  0  +  j  sin  0  =  -0.9992  +  jO.0414 
cos  0  =  -0.9991  and  sin  0  =  0.0414 

S  =  +177.6  degrees. 
A  vector  diagram  for  equation  (26)  is  shown  in  Fig.  5,  in  which 

,50  (0.866  +  j  0.600) 


V 6026(1+ JO)  V-5965U+JO) 

.   FIG.  5. 

the  vectors  corresponding  to  the  negative  solution  are  shown 


VECTORS  AND  COMPLEX  QUANTITIES  15 

dotted.  The  angles  and  vectors  in  Fig.  5  are  not  drawn  exactly 
to  scale.  Their  relative  magnitudes  are  purposely  altered  to 
make  the  diagram  clearer. 

Exponential  Operator  e±J$.  —  When  vectors  are  to  be  added 
or  subtracted,  the  operator  (cos  a  ±  j  sin  a)  should  be  used  with 
each  in  order  that  the  vectors  may  have  the  proper  phase  rela- 
tions. Vectors  when  expressed  in  the  complex  form,  i.e.,  in  the 
form  a  +  jb,  may  be  added  or  subtracted,  multiplied  or  divided. 
When,  however,  the  operations  of  multiplication  and  division 
only  are  to  be  performed,  the  exponential  operator  e±'e  is  much 
more  convenient  and  should  ordinarily  be  used.  The  exponential 
operator  is  particularly  convenient  in  such  cases  since  the  pro- 
cesses of  multiplication  and  division  then  involve  only  the  addi- 
tion and  subtraction  of  exponents.  The  exponential  form  of 
operator  should  not  be  used  when  vectors  are  to  be  added  or 
subtracted,  since  these  operations  cannot  be  performed  by  the 
mere  addition  and  subtraction  of  exponents. 
The  expansion  of  e>e  by  Maclaurin's  theorem*  gives 

, 

- 


02   ,    04       6s    . 

-  {2  +  0  ~  j  +  etc- 

03         05         f)~ 

+^-|  +  ^-|  +  etc-)  (28) 

where,  for  example,  |5  means  1X2X3X4X5  and  e  is  the 
base  of  the  Napierian  logarithms,  i.e.,  2.718. 

Similarly,  the  expansion  of  the  sine  and  cosine  of  0  gives 

sin  0   =  0  -|  +  ^  -  ~  +  etc.  (29) 

f)2       ft*       #6 
cos  0  =  1  -    ^  +  ^  -  •  ^  +  etc.  (30) 

Therefore 

€»'*  =  cos  0  +  j  sin  0  (31) 

*  See  any  standard  book  on  calculus. 


16 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


The  expansion  of  e~]'e  gives 


I 


+ 


-f  etc. 


/    — 

2  n  '13 

r  4 

e-  + 

2 

04       06 

4       |6H 

-  etc. 

—  j 

•(*-?H 

05   _ 

—  etc. 


etc.  ) 


=  cos  0  —  j  sin  0 


(32) 


eie  and  e~'*  are  therefore  two  operators  producing  on  a  vector 
to  which  they  are  applied  the  same  effect  as  the  operators 
(cos  6  +  j  sin  0)  and  (cos  6  —  j  sin  6)  .  e>e  is  an  operator  which 
rotates  a  vector  to  which  it  is  applied  through  an  angle  6  in  a 
positive  direction.  e~J'0  is  an  operator  which  rotates  a  vector  to 
which  it  is  applied  through  an  angle  0  in  a  negative  direction. 
The  angle  6  in  the  two  operators  should  be  expressed  in  radians 
to  be  mathematically  correct,  but  it  is  usually  more  convenient 
to  express  it  in  degrees. 

A  (cos  0  +j  sin  6)  =  Ae*e  (33) 

A  (cos  e  -  j  sin  0)  =  Ae-"  (34) 

represent  two  vectors  which  make 
angles  of  plus  6  and  minus  0  with  the 
axis  of  reals. 

The  operator  (cos  0  ±  j  sin  0)  refers  a 
vector  to  which  it  is  applied  to  rectan- 
gular coordinates.  The  operator  e±30  re- 
fers a  vector  to  polar  coordinates.  The 
latter  operator  is  called  the  exponential 
operator.  The  vectors  given  in  equations 
(33)  and  (34)  are  shown  in  Fig.  6. 

Let  x  =  (a  +  jb)  and  y  =  (c  +  jd)  be  two  vectors  making 
angles  6X  =  tan"1-  and  6y  =  tan"1  with  some  common  reference 

d  C 

axis.     Then 

xy  -  (o  +  jb)  (c  +  jd) 

=  xy(cos  6X  +  j  sin  Ox)  (cos  6 
=  xy\cos(0x  +  ey)  +  j  sin(0x 

(35) 


FIG.  6. 


j  sin  0y) 
8,)} 


VECTORS  AND  COMPLEX  QUANTITIES  17 

Similarly  if  q  =  q(cos  0q  +  j  sin  0q)  and  z  =  z  (cos  0,  —  j  sin  02) 
are  two  other  vectors 

qz  =  qz  (cos  0g  +  j  sin  0g)  (cos  0Z  —  j  sin  0Z) 
=  qz{cos(0q  -  02)  +  j  sin(0,  -  0,)) 
=  qzej  (e"  ~  fc)  (36) 

x  =  a+jb  =  /x\  /cos  0x+jsin0A 
y~c+jd       \y)  Vcos  0y  +  j  sin  0y/ 
_  a^  /cos  0X  +  j  sin  0A  /cos0v  —  jsin0y\ 
y  Vcos  0tf  +  j  sin  0V7  Vcos  0y  —  j  sin  dj 
=  x  /cos(0a  -  0V)  +  j  sin(0z  -  0y)\ 
y  \  cos20v  +  sin20v  / 

=  -{cos(0z  -  ey)  +  j  sin(0x  -  0y)} 
y 

-  *J(9f  ~  •.) 

/  (37) 

An  Example  of  the  Use  of  the  Exponential  Operator  e±J0.— 

Suppose  that  the  product  of  three  vectors,  of  magnitudes  10, 
15  and  20  making  angles  of  +  15,  +  20  and  —  120  degrees 
respectively  with  a  fixed  reference  axis,  is  to  be  divided  by  a 
fourth  vector  of  magnitude  10  making  an  angle  of  +  10  degrees 
with  the  same  reference  axis.  The  four  vectors  are 


10{cos   15°  +  j  sin   15°  }  = 
15  1  cos  20°  +j  sin  20°  }  =  15c>200 
20{cos(-  120°)  +  jsin(  -  120°)}  =  20e-J'120° 
10  {cos    10°  +j  sin  10°  }  =  ICe'10" 

To  be  mathematically  correct  the  angles  in  the  exponents  of 
the  e's  should  be  in  radians.  It  is,  however,  more  convenient 
to  leave  them  in  degrees. 

The  result  of  the  operation  of  multiplication  and  division, 
using  the  exponential  operators,  is 

(  10  X  15  X  20]  f€»'lfi°  X  e>20°  X  e->120° 
A  ~ 


10  6'10' 

=  300  e^15"  +  20°  ~  120°  "  100) 
=  300e-'95° 

=  300 (cos  95°  -  j  sin  95°) 
=  300(- 0.0872  -  J0.9962) 
=  -  26.16  -  J298.8 


18  PRINCIPLES  OF  ALTERNATING  CURRENTS 

If  the  result  had  been  obtained  by  using  the  operator 
(cos  6  ±  j  -sin  6)  for  rectangular  coordinates,  it  would  first  have 
been  necessary  to  rationalize  the  expression.  There  would  then 
have  been  four  quantities  of  the  form  (a  ±  jb)  to  multiply  to- 
gether in  the  numerator  and  two  to  multiply  together  in  the 
denominator.  To  get  the  result  in  the  final  form  of  (a  ±  jb), 
the  real  terms  in  the  numerator  would  then  have  to  be  added  and 
divided  by  the  denominator  to  get  the  real  or  a  part  of  the 
resultant  vector  and  the  imaginary  terms  in  the  numerator  would 
have  to  be  added  and  divided  by  the  denominator  to  get  the 
imaginary  or  b  part  of  the  resultant  vector.  The  saving  of  time 
by  the  use  of  the  exponential  operator  is  obvious  in  a  case  of  this 
kind. 

Exponential  Operator  which  Produces  Uniform  Rotation.— 
If  the  angle  6  in  the  operator  e±<7°  is  replaced  by  an  angle  which 
is  proportional  to  time,  the  operator  will  produce  continuous 
rotation  of  any  vector  to  which  it  is  applied.  Let  0  =  ut,  where 
co  is  an  angular  velocity  of  2irf  radians  per  second  and  t  is  the  time 
measured  in  seconds.  Then  e±jut  is  an  operator  producing 
a  continuous  rotation  of  2irf  radians  or  /  revolutions  per  second. 

A     =Ae+jut  (38) 

A'  =  Afe~iut  (39) 

are  two  rotating  vectors  which  have  magnitudes  A  and  A'  and 
rotate  with  a  uniform  angular  velocity  of  2wf  =  co  radians  per 
second.  They  make  /  revolutions  per  second.  A  revolves  in  a 
positive  or  counter-clockwise  direction.  .  A'  revolves  in  a  nega- 
tive or  clockwise  direction. 

B  =  Bejut  (40) 

C  =  Cei(ut~e)  (41) 

are  two  rotating  vectors  having  magnitudes  of  B  and  C  and 
revolving  in  a  counter-clockwise  direction  with  an  angular  velo- 
city of  co  =  27r/  radians  per  second.  C  lags  behind  B  by  6 
radians.  These  continuously  rotating  vectors  may  be  expressed 
in  terms  of  rectangular  coordinates  instead  of  in  polar  coordinates 
by  the  use  of  the  operator  (cos  co£  ±  j  sin  co£).  (See  page  12.) 

B  =  B<iut  =  £jcos  to*  +  j  sin  cot]  (42) 

C=  C€j(at~e)  =  CjcosM  ~  9)  +  j  sinM  -  0)}          (43) 


VECTORS  AND  COMPLEX  QUANTITIES 


19 


Polar  Form  of  Operator. — It  is  often  convenient  to  represent 
a  vector  by  writing  its  magnitude  followed  by  a  phase  angle 
which  gives  its  phase  position  with  respect  to  the  reference  axis. 
For  example: 

Z  =  A  9  (44) 

A'=A\^0  (45) 

are  two  vectors  of  the  same  magnitude  A,  making  angles  of 
+  6  and  —  0  respectively  with  the  reference  axis.  The  two 
vectors  A\Q_  and  A  —  0  are  shown  in 
Fig.  7. 

|+0  =  |0  and  |~0  =  1 0  are  symbolic  oper- 
ators which  indicate  that  the  vectors  to 
which  they  are  attached  have  been  ro- 
tated through  angles  of  +  6  and  —  6  re- 
spectively. Instead  of  using  the  sign 
with  the  angle  to  indicate  whether  it  is 
positive  or  negative,  the  position  of  the 
bracket  may  be  made  to  serve  this  purpose.  When  this  is 
done  |0  means  a  positive  angle  and  \B  means  a  negative  angle. 
The  polar  form  of  operator  is  simpler  than  the  exponential 
form,  is  just  as  satisfactory  and  for  this  reason  is  more  often  used. 

Example  of  the  Use  of  the  Polar  Form  of  Operator. — Suppose 
it  is  necessary  to  divide  the  product  of  three  vectors,  of  magnitudes 
A,  B,  and  C  making  respectively  angles  of  BA,  —On  and  Bc  with 
some  reference  axis,  by  the  product  of  two  other  vectors  of 
magnitudes  D  and  E  making  angles  respectively  of  —  6D  and 
BE  with  the  same  axis.  The  vector  representing  the  result  of  this 
operation  is 


FIG.  7. 


(D  -OD 


'(BA  —  BB  +  BC)  —  ( —  BD  +  BE) 

[COS  (BA  -  BB  +  Bc  +  BD  -  BE) 

j  sin  (0A  ~  BB  +  Bc  +  0D  -  0^)  1 


20  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Example  of  the  Solution  of  a  Vector  Equation  having  Two 
Terms,  Each  being  Expressed  by  Means  of  the  Polar  Form  of 
Operator. — The  current  in  a  certain  transmission  line  under  load 
conditions  is  given  by  the  following  vector  equation  in  which  the 
voltage  at  the  receiving  end  is  used  as  the  axis  of  reference, 
i.e.,  as  the  axis  of  reals. 

I  =  175  -25°84  X  0.9919  |  +0?313 

+  63,510   +0°0  X  0.002543    +15?38  X  0.1386  |  +74°72 
=  175X0.9919  -25°84  +  0?313 


+  63,510  X  0.002543  X  0.1386  |+0°0  +15°38  +74°72 
=  173.6    -25?53  +  22.38  |  +90?10 

=  173.6  (cos  25°53  -  j  sin  25?53) 

+  22.38  (cos  90°10  +  j  sin  90°10) 

=  173.6  (0.9024  -  jO.4310)  +  22.38  (-0.0017  +  jl.OOO) 
=  156.6  -  J52.43 


I  =       (156.6)2  +  (52.43)2 
=  165.2  amperes. 


e  =  -is°5i 

I  =  165.2|^18°51  =  165.2 
=  165.26-  ''18?51 


Square  Root,  Product  and  Ratio  of  Vectors  or  of  Complex 
Quantities  by  Use  of  the  Exponential  Operator.  —  A  method  of 
getting  the  root  or  power  of  a  vector  or  complex  quantity  by  the 
use  of  the  operator  (cos  0  ±  j  sin  6)  has  been  given  on  page  9. 
Roots  and  powers  may  be  obtained  much  more  easily  by  using 
the  exponential  operator.  Let 

A  =  a  +  jb 

be  any  vector  making  an  angle  a  with  the  axis  of  reference.     Then 

A  =  A<ia 


VECTORS  AND  COMPLEX  QUANTITIES  21 


A  vector  is  not  altered  in  direction  or  magnitude  by  multiplying 
it  by  an  operator  which  rotates  it  through  any  whole  number  of  2ir 
radians,  i.e.,  by  multiplying  it  by  the  operator  e*2*5  where  q  is 
any  integer.  Therefore 


A  =  AS"  =  AJ"     *q  =  A<i 

(1+T9)  (46) 


There  are  only  two  different  roots  of  the  vector  A.     There 
are  two  roots  of  \^A  ,  one  plus  and  one  minus.     There  are  also 

two  roots  of  the  operator  e^*2    *q'  .     These  are  for  q  =  0  and 
5  =  1.     For  values  of  q  greater  than  1  the  roots  repeat. 

a 

For  q  =  0,  \/?"  =  e  '2 

./a  .      \ 

For  q  =  1,  V?°  =  • 


Adding  TT  to  the  exponent  of  the  operator  is  equivalent  to 
reversing  its  sign.     The  two  roots  of  the  operator  are  therefore 


The  two  roots  of  the  vector  are 


±VA  £  =    ±  VA  (cos  \  +  j  sin  ^)      (47) 


The  square  root  of  a  vector  has  therefore  two  values.  It  is  a 
new  vector  having  a  magnitude  equal  to  the  square  root  of  the 
magnitude  of  the  original  vector  and  a  phase  angle  equal  either 
to  half  the  phase  angle  of  the  original  vector  or  to  half  the  phase 
angle  of  the  original  vector  plus  IT. 

The  two  values  of  the  square  root  of  a  vector  are  shown  in 
Fig.  8. 


22 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Let  B  =  B^  be  a  second  vector  of  magnitude  B  making  an 
angle  0  with  the  same  reference  axis  that  was  used  for  the  vector 
A.  Then 

AB  =  Aet*B<P 

=  AB<?(a  +  fl>  =  AB{cos  (a  +  ft  +  j  sin  («  +  ftj     (48) 

A  =  Ae  30i  m  A  (cos  a  +  j  sina) 


Z  {cos(    +  TT)  +  j  sin 


FIG.  8. 


The  product  of  two  vectors  is  a  new  vector  having  a  magnitude 
equal  to  the  product  of  the  magnitudes  of  the  original  vectors. 
It  makes  an  angle  with  the  reference  axis  equal  to  the  sum  of  the 
phase  angles  of  the  original  vectors.  This  was  shown  on  page  9, 
but  less  simply,  by  the  use  of  the  operator  (cos  6  +  j  sin  8)  . 


A  A  fja  A  A 

=          =       ("  ~ft}  =      icos  (a  ~ 


j  sin  (a  ~ 


(49) 


The  ratio  of  two  vectors  is  a  new  vector  having  a  magnitude 
equal  to  the  ratio  of  the  magnitudes  of  the  original  vectors.  It 
is  displaced  from  the  reference  axis  by  an  angle  equal  to  the 
difference  of  the  phase  angles  of  the  original  vectors. 


(50) 


The  reciprocal  of  a  vector  is  a  new  vector  having  a  magnitude 
equal  to  the  reciprocal  of  the  magnitude  of  the  original  vector. 


VECTORS  AND  COMPLEX  QUANTITIES  23 

It  is  displaced  from  the  reference  axis  by  an  angle  equal  in  magni- 
tude but  opposite  in  direction  to  the  phase  angle  of  the  original 
vector. 


VAB  =  ± 

=  ±VAB  |  cos  (^-^)  +  i  efa(^jfy         (5i) 

i  =  ±v'I/(^) 


The  fundamental  constants  which  appear  in  the  exact  solution 
of  the  transmission  line  for  current  and  voltage  are  the  square 
root  of  the  product  and  the  square  root  of  the  ratio  of  the  series 
impedance  per  unit  length  of  conductor  and  the  parallel  admit- 
tance to  neutral,  also  per  unit  length  of  conductor.  Impedance 
gj)H  fulmi±tfl.nfift  nrt*  not  vectors.  They  are  complex,  quantities. 
They  are  really  special  operators.  If  a  current  vector  is  operated 
on  with  impedance,  the  result  is  a  voltage  vector  which  differs 
both  in  phase  and  in  magnitude  from  the  current  vector.  Simi- 
larly if  a  voltage  vector  is  operated  on  with  admittance,  the  result 
is  a  current  vector  which  differs  both  in  magnitude  and  in  phase 
from  the  voltage  vector.  So  far  as  concerns  the  purely  mechani- 
cal processes  of  multiplication,  division,  extraction  of  roots,  etc. 
of  complex  quantities,  they  are  carried  out  in  the  same  manner 
for  complex  quantities  as  for  vectors.  For  the  solution  of  the 
equations_fpj^the_Jrajnsjnission  line  it__is  necessary  to  be  able 
to  take  the  square  root  of  the  product  and  the  square  root  of_  the 
ratio  of  complex  quantities. 

Complex  Quantity.  —  A  complex  quantity  is  a  quantity  which 
multiplied  by  a  vector  gives  a  new  vector  of  a  different  kind  and 
magnitude  from  the  original  vector  and  displaced  from  it  in 
phase.  A  complex  quantity  is  a  simple  algebraic  quantity 
combined  with  an  operator.  For  example,  the  inductive  impe- 
dance of  an  electric  circuit  is  a  complex  quantity.  It  is  written 

r  +jx  =  z<ie  =  z  (cos  6  +  j  sin  6)  I 
where  the  letters  have  the  following  significance. 


24  PRINCIPLES  OF  ALTERNATING  CURRENTS 

r  =  resistance 

X   =   27T/L 

/  =  frequency 
L  =  inductance 
2  =  \/r2  +  x2 

j  =  V  —  1  =  operator  which  rotates  through +90  degrees 
c  =  base  of  the  Napierian  logarithms 

6  =  tan-1  - 

jmpedance  multipliejj_by._current  gives  voltage.  This  voltage 
is  displaced  from  the  current  by  an  angle  9. 

In  general  the  negative  root  obtained  when  taking  the  square 
root  of  a  complex  quantity,  such  as  impedance,  has  no  significance 
and  does  not  have  to  be  considered. 

The  nth  Root  and  nth  Power  of  Vectors  and  Complex  Quanti- 
ties.— In  ordinary  electrical  engineering  problems  it  is  not  often 
necessary  to  use  roots  or  powers  of  vectors  or  complex  quantities 
higher  than  the  second.  Higher  roots  and  powers  can  be  ob- 
tained, however,  if  necessary.  A  method  for  obtaining  them 
using  the  operator  (cos  0  +  j  sin  6)  has  already  been  given. 
They  may  be  found  more  easily  by  using  the  operator  eje.  For 
example,  if  A  =  Aeja: 

/a  +  2vq\ 


This  will  have  n  different  roots. 

An  =  (Ae'T  =  AV™  =  An(cos  no.  +  j  sin  no)          (54) 
(AB)n  =  (Ae**B*)n  =  AnBn(/n(a  +  ()) 

=  AnBn{cos  n(a  +  0) 

+  jsinn(a  +  0)}    (55) 


'«    n     An  y»(«-« 


~  {cos  n(a-ff)  +  j  sin  n(a  -  0)  j        (56) 


VECTORS  AND  COMPLEX  QUANTITIES  25 

Logarithm  of  a  Complex  Quantity  or  a  Vector. — Let  A  =  a  + 
jb  be  any  complex  quantity  or  vector.  This  may  be  written 

A  =  a  +  jb  =  A  (cos  a  -\-  j  sin  a)  =  At*" 

where  «  =   tan"1  - 
a 

log,  A  =  log,  Ae*" 

=  log,  A  +  j  a  log,€ 
=  log,  A  +  ja 

=  log,  Va2  +  b2  +  j  (fcwr"1^)  (57) 

The  logarithm  of  a  vector  or  a  complex  quantity  is  a  new  vector 
or  complex  quantity  having  a  real  part  equal  to  the  logarithm 
of  the  magnitude  of  the  original  vector  or  complex  quantity 
and  an  imaginary  part  equal  to  the  phase  angle;  expressed  in 
radians,  of  the  original  vector  or  complex  quantity. 

Representation  of  an  Oscillating  Vector,  whose  Magnitude 
varies  Sinusoidally  with  Time,  by  the  use  of  Two  Exponential 
Rotating  Operators. — The  motion  of  the  end  of  a  spiral  spring 
that  has  been  compressed  and  released  is  an  example  of  an  oscil- 
lating vector  whose  magnitude  varies  sinusoidally  with  time. 
A  sinusoidal  alternating  current,  i.e.,  one  whose  magnitude 
varies  sinusoidally  with 
time,  may  be  represented 
by  a  similar  vector. 

Any  simple  harmonic 
oscillating  vector  may  be 
resolved  into  two  op- 
positely rotating  vectors, 
each  of  the  same  period 
as  the  given  vector  and  of  FIG.  9. 

half  its  magnitude. 

In  Fig.  9,  A  is  a  vector,  of  maximum  value  A,  which  oscillates 
in  magnitude  between  the  limits  of  +A  and  —A.  The  magni- 
tude of  this  vector  at  each  instant  is  given  by 

a  =  A  sin  (orf  +  90°)  =  A  cos  wt 

where  co  is  equal  to  2ir  times  the  frequency  of  vibration  of  the 
vector  A .  It  is  also  equal  to  the  angular  velocity  of  the  rotating 
vectors  which  replace  A .  A  i  and  A  2,  each  equal  in  magnitude  to 


26  PRINCIPLES  OF  ALTERNATING  CURRENTS 

A 

«-,  are  the  two  oppositely  rotating  vectors  which  replace  the 

vector  A.     Their  resultant  always  lies  along  the  original  vector 
and  is  equal  to  the  magnitude  of  the  vector  at  each  instant. 

Referring  the  two  rotating  vectors  to  +A  as  a  reference  axis 
gives 

T!  =  ^  (cos  ^t  +  j  sin  «0  =  ^  e+^ 

A  A 

A  2  =  -«  (cos  co£  —  j  sin  orf)  =  -^  e~jat 

a  =  A  cos  at  =  At  +  ^2  =  £  €*•*  +  4  «"**  (58) 

z  ^ 

a'  =  A  sin  «*  =  4  €+y("*-0  +  4  «''("""*)  (59) 

.i  — 

If  ii  is  the  instantaneous  value  of  a  sinusoidal  alternating 
current  whose  maximum  value  is  Imi, 

ii  =  /«i  sin  at  =  ^  e^'C^-D  +  ^  €-y(^-0          (60) 

In  this  expression  ii  is  zero  when  time,  t,  is  zero. 

If  i%  is  another  sinusoidal  alternating  current  of  the  same 
frequency  as  ii,  but  lagging  it  in  time  phase  by  an  angle  0, 

t,  =  /.,  sin  («<  -  «)  =  *-f  «+<--l-0  +  ^-2  e-<—2-')     (16) 

This  method  of  representing  an  alternating  current  is  cumber- 
some and  has  little  to  recommend  it  under  ordinary  conditions. 

Representation  of  a  Vector  which  Rotates  with  an  Uniform 
Angular  Velocity  and  Shrinks  in  Magnitude  Logarithmically  with 
Time.— 

x  =  Aea<(cos  oil  +  j  sin  co£) 


is  a  revolving  vector  which  rotates  with  an  uniform  angular 
velocity  cot  and  shrinks  logarithmically  with  time  at  a  rate 
determined  by  the  constant  exponent  a.  A  is  a  constant  and 
is  equal  to  the  length  of  the  vector  when  time,  t,  is  zero. 

If  the  angle  ut  is  measured  from  the  horizontal  axis,  the  projec- 
tion on  a  vertical  axis  of  the  revolving  vector  x  =  Aeate±J'at  will 
represent  a  sinusoidally  oscillating  quantity  whose  successive 
maximum  values  decrease  logarithmically  with  time.  The 
current  produced  by  the  discharge  of  a  condenser  through  an 
inductive  circuit  of  low  resistance  may  be  represented  in  this  way. 


CHAPTER  II 
ALTERNATING  CURRENTS 

£>irect  Current  or  Voltage. — A  direct  current  or  voltage 
is  unidirectional,  but  it  may  pulsate.  However,  the  terms 
direct  current  and  direct  voltage,  as  ordinarily  used,  designate 
currents  and  voltages  which  are  practically  steady  and  non- 
pulsating. 

Pulsating  Current  or  Voltage. — A  pulsating  current  or  pulsat- 
ing voltage  is  one  that  pulsates  regularly  in  magnitude.  The 
term  pulsating  as  ordinarily  applied  to  a  current  or  to  a  voltage 
rftfers  to  a  unidirectional  current  or  voltage. 

Continuous  Current  or  Voltage. — A  continuous  current  or 
voltage  is  one  which  is  practically  non-pulsating. 

Alternating  Current  or  Voltage. — An  alternating  current  or 
voltage  is  one  that  alternates  regularly  in  direction.  The  term 
when  applied  to  a  current  or  voltage  usually  refers  to  a  current 
or  voltage  which  is  periodic  and  whose  successive  half  waves  are 
of  the  same  shape  and  area. 

Oscillating  Current. — An  oscillating  current  is  a  periodic 
current  whose  frequency  is  determined  by  the  constants  of  the 
circuit  in  which  it  is  produced. 

Instantaneous  Value. — The  instantaneous  value  of  a  current 
or  voltage  is  its  value  at  any 
given  instant  of  time  t. 

Cycle. — A  cycle  of  an  alter- 
nating current  or  voltage  is  one 
complete  set  of  positive  and 
negative  values  of  the  current 
or  voltage.  These  values  re- 
peat themselves  at  regular 
intervals. 

Periodic  Time  or  Period. — The  periodic  time  or  period  of  an 
alternating  current  or  voltage  is  the  time  required  for  it  to  pass 
through  one  complete  cycle  of  values.  It  is  expressed  in  seconds 
and  is  ordinarily  denoted  by  T.  A  cycle  is  represented  in  Fig. 
10  by  the  portion  of  the  wave  between  a  and  b. 

27 


28  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Frequency. — The  number  of  cycles  passed  through  by  an 
alternating  current  or  voltage  in  a  second  is  its  frequency.  It  is 
denoted  by  the  letter  /.  For  example,  a  60-cycle  circuit  is 
one  in  which  the  current  and  voltage  pass  through  60  complete 
cycles  per  second.  Its  frequency  and  periodic  time  would  be, 

respectively,  /  =  60  cycles  and  T  =  j  =  go  =  0.01667  second. 

The  frequencies  commonly  used  in  America  are  60  and  25  cycles, 
although  50  cycles  is  also  used.  Abroad,  50  cycles  is  common  and 
there  are  a  few  installations  for  railway  work  with  frequencies 
lower  than  25  cycles.  In  general,  frequencies  as  low  as  25  cycles 
are  unsatisfactory  for  lighting,  on  account  of  the  noticeable 
flicker  in  the  lights  produced  in  many  cases  by  so  low  a  frequency. 
Sixty  cycles  is  almost  universally  used  for  lighting,  although  50 
cycles  is  perfectly  satisfactory.  Twenty-five  cycles  was  formerly 
used  for  long-distance  power  transmission  and  for  installations 
where  power  was  the  primary  object.  However,  due  to  the 
better  design  of  apparatus  and  also  to  a  better  understanding 
of  the  problems  of  long-distance  power  transmission,  sixty 
cycles  is  now  largely  being  used  for  long-distance  power  trans- 
mission as  well  as  for  ordinary  power  and  lighting.  In  general, 
60-cycle  apparatus  is  lighter  and  cheaper  than  apparatus  built 
for  25  cycles.  The  difference  in  weight  and  cost  between  60-cycle 
and  25-cycle  apparatus  is  essentially  the  same  as  between  high- 
and  low-speed  direct-current  motors  and  generators. 

The  preceding  definitions  for  the  most  part  are  taken  with 
slight  modifications  from  the  revised  (1921)  Standardization 
Rules  of  the  American  Institute  of  Electrical  Engineers. 

Wave  Shape  or  Wave  Form. — The  shape  of  the  curve  obtained 
when  the  instantaneous  values  of  a  voltage  or  current  are  plotted 
as  ordinates  against  time  as  abscissas  is  its  wave  shape  or  wave 
form.  The  abscissa  for  one  cycle  is  taken  as  2ir  radians  or  360 
degrees,  and  would  correspond  to  one  complete  revolution  of  the 
armature  of  a  two-pole  alternator. 

Two  alternating  currents  or  voltages  are  said  to  have  the  same 
wave  form,  when  their  ordinates,  at  corresponding  positions  in 
degrees  measured  from  the  zero  points  of  the  waves,  bear  a 
constant  ratio  to  each  other. 


ALTERNATING  CURRENTS  29 

Simple  Harmonic  Current  or  Voltage. — The  form  of  periodic 
current  or  voltage  most  easily  dealt  with  mathematically  is 
one  whose  instantaneous  values  follow  one  another  according  to 
the  law  of  sines.  In  general  this  wave  form  is  most  desirable 
from  the  standpoint  of  the  generation,  transmission  and  utili- 
zation of  power.  The  sine-wave  form  of  a  periodic  voltage, 
i.e.,  a  sinusoidal  voltage,  is  given  by  equation  (1),  where  the  time, 
t,  is  reckoned  from  the  instant  when  the  voltage  is  zero  and  in- 
creasing in  a  positive  direction. 

v  =  Vm  sin  27r  t=  vm  sin  2irft  =  Vm  sin  at  (1) 

Vm  is  the  maximum  value  of  the  wave  or  its  amplitude.  T 
is  the  duration  of  one  complete  cycle,  i.e.,  the  periodic  time,  and 
/  is  the  number  of  complete  cycles  per  second,  i.e.,  the  frequency. 

A  simple  harmonic  or  sinusoidal  wave  is  positive  during  one 
half  of  each  cycle  and  negative  during  the  other  half  of  the  cycle. 
A  simple  harmonic  current  is  plotted  in  Fig.  10  with  angles  as 
abscissas.  Fig.  10  is  its  wave  form.  The  abscissas  might  equally 
well  have  been  time,  2ir  radians  corresponding  to  the  time  of 

one  complete  cycle,  i.e.,  to  T  =  j  second.     The  abscissas  are 

indicated  in  both  these  ways  in  the  figure. 

Alternating  Current  Calculations  Based  on  Sine  Waves. — 
Alternating-current  calculations  are  commonly  based  on  the 
assumption  of  sinusoidal  waves  of  current  and  voltage.  Where 
the  waves  are  not  sinusoidal,  it  is  possible  to  resolve  them  into 
component  sinusoidal  waves,  known  as  the  fundamental  and 
harmonics,  having  frequencies  which  are  1,  2,  3,  4,  5,  6,  etc. 
times  the  frequency  of  the  circuit.  The  components  which 
have  frequencies  equal  to  2,  4,  6,  etc.  times  the  frequency  of  the 
circuit,  i.e.,  the  so-called  even  harmonics,  are  not  present  as  a 
rule  in  waves  of  current  or  voltage.  (See  Chapter  IV.)  The 
effects  of  the  different  component  sinusoidal  waves  may  be 
determined  separately  and  then  combined. 

Since  any  periodic  alternating  current  can  be  analyzed  into  a 
series  of  simple  sinusoidal  waves  of  definite  frequencies,  it  is 
desirable  to  discuss  in  considerable  detail  the  conditions  which 
hold  for  circuits  carrying  simple  sinusoidal  or  simple  harmonic 
currents. 


30  PRINCIPLES  OF  ALTERNATING  CURRENTS 

A  periodic  current  or  voltage  need  not  follow  the  simple  sine 
law,  but  its  instantaneous  values  must  be  some  function  of  time 

and  it  must  go  through  a  complete  cycle  in  -j  second.    Its  positive 

and  negative  loops  need  not 
be  symmetrical,  but  on  ac- 
count of  the  way  in  which 
periodic  waves  are  generated, 
the  positive  and  negative 
tm^  *~T  loops  are  symmetrical  with 

respect  to  the  axis  of  time, 
except  in  very  rare  cases. 
A  periodic  current  which 
does  not  follow  the  simple 
sine  law  is  shown  in  Fig.  11. 

Phase. — Time  need  not  be  reckoned  from  the  instant  when  a 
current  or  voltage  wave  passes  through  zero  and  in  many  cases 
it  is  not  so  reckoned.  If  time  is  considered  zero  at  a  point  a 

radians  (equivalent  to  ^-  X  7    second)  after  the  wave  has  passed 
Zir     j . 

through  zero  increasing  in  a  positive  direction,   equation   (1) 
becomes 

v  =  Vm    sin(co£  +  a)  (2) 

The  angle  a.  is  known  as  the  phase  angle  of  the  wave.  It 
indicates  the  number  of  radians  between  the  point  where  the 
wave  is  zero  and  increasing  in  a  positive  direction  and  the  point 
from  which  time  is  reckoned.  The  angle  may  be  either  positive 
or  negative.  When  positive  it  is  an  angle  of  lead  and  the 
voltage  goes  through  zero  increasing  in  a  positive  direction  before 
time,  t,  is  zero.  When  the  angle  is  negative  it  is  an  angle  of  lag 
and  the  voltage  goes  through  zero  increasing  in  a  positive  direction 
after  time,  t,  is  zero. 

Equations  (3)  and  (4)  represent  a  voltage  and  a  current  of  the 
same  frequency,  both  of  which  are  sinusoidal  but  which  do  not 
go  through  their  zero  values  at  the  same  instant. 

v  =  Vm  sin(orf  +  ai)  (3) 

i  =  Im  sin(orf  +  «2)  (4) 


ALTERNATING  CURRENTS  31 

Although  both  waves  have  the  same  frequency,  they  are  not  in 
phase.  Their  difference  in  phase  is  the  difference  between  their 
phase  angles.  When  the  phase  angles  are  angles  of  lead  as  in 
equations  (3)  and  (4),  the  angle  of  lead  of  v  with  respect  to  i 
is  the  phase  angle  of  v  minus  the  phase  angle  of  i  or  (ai  —  «2). 
The  angle  of  lead  of  i  with  respect  to  v  is  the  phase  angle  of 
i  minus  the  phase  angle  of  v  or  («2  —  «i).  The  corresponding 
angles  of  lag  are  (a2  —  «i)  for  v  with  respect  to  i  and  («i  —  «2) 
for  i  with  respect  to  v.  When  «2  is  less  than  «i,  («i  —  «2)  is 
positive  and  v  actually  leads  i.  When  «2  is  greater  than  a\, 
(«i  —  «2)  is  negative  and  («2  —  «i)  is  positive.  In  this  case 
v  leads  i  by  a  negative  angle  and  lags  i  by  a  positive  angle. 
Since  a  negative  angle  of  lead  is  equivalent  to  a  positive  angle 
of  lag,  when  (a\  —  a*)  is  negative  v  actually  lags  i.  A  negative 
angle  of  lead  is  always  equivalent  to  an  actual  angle  of  lag,  and, 
similarly,  a  negative  angle  of  lag  is  always  equivalent  to  an  actual 
angle  of  lead. 

Although  the  phase  angles  in  equations  (3)  and  (4)  should  be 
expressed  in  radians  to  be  mathematically  correct,  it  is  usually 
more  convenient  to  express  them  in  degrees.  When  this  is  done, 
both  wt  and  a  must  be  reduced  to  the  same  units,  i.e.,  radians 
or  degrees,  before  they  can  be  added. 

Generation  of  an  Alternating  Electromotive  Force. — The 
electromotive  force  generated  or  induced  in  a  coil  through  which 
flux  is  varying  is  equal  to  the  rate  of  change  of  flux  through  the 
coil  with  respect  to  time  multiplied  by  the  number  of  turns  in  the 
coil,  that  is,  it  is  equal  to  the  time  rate  of  change  of  flux  linkages 
for  the  coil.  The  flux  linkages  are  equal  to  the  flux  through  the 
coil  multiplied  by  the  number  of  turns  with  which  this  flux  links. 
If  e  represents  the  instantaneous  voltage  induced  in  a  coil  by  a 
change  in  the  flux  linking  it 

•  =  -  4:  (5) 

The  time  rate  of  change  of  flux  through  the  coil  is  -ri  and  N  -r. 

is  the  time  rate  of  change  of  flux  linkages  with  the  coil.  The 
voltage  e  in  equation  (5)  is  a  voltage  rise.  When  the  minus  sign 
is  omitted  it  is  a  voltage  fall  or  drop.  (For  the  significance  of  rise 
and  fall  applied  to  a  voltage  see  page  64.) 


32  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  electromotive  forces  induced  or  generated  in  the  armature 
turns  of  a  direct-current  generator  are  alternating  but  they  are 
rectified  with  respect  to  the  external  circuit  by  means  of  the 
commutator. 

Figure  12  shows  a  simple  two-pole  alternator  with  a  revolving 
armature  and  a  single  armature  coil. 

N  and  S  are  the  poles.  A  is  the  armature  which  carries  a 
winding  placed  in  two  diametrical  slots  a  and  6.  The  terminals  of 

this  winding,  in  an  actual  machine, 
would  be  brought  out  to  two  insulated 
slip  rings  mounted  on  the  armature 
shaft.  The  current  would  be  taken 
from  these  slip  rings  by  means  of 
brushes.  If  the  poles  of  the  alternator 
are  so  shaped  that  the  flux  through 
the  coil  varies  as  the  cosine  of  the 

angular  displacement  of  the  coil  from  the  vertical  position  (see 
Fig.  12)  a  sinusoidal  wave  of  electromotive  force  will  be 
generated.  In  this  case 

e  =  ~N%  =-**&*-<*»<* 

=  coN(pm  sin  ut  (6) 

N  is  the  number  of  turns  in  the  armature  winding  and  co 
is  the  angular  velocity  of  the  armature  in  radians  per  second. 
The  maximum  value  of  the  flux  through  the  coil  is  <pm.  In 
case  the  alternator  has  more  than  two  poles  there  will  be  as 
many  similar  winding  elements  as  there  are  pairs  of  poles. 
These  will  be  spaced  at  a  distance  apart  on  the  armature  equal  to 
the  distance  between  two  adjacent  poles  of  like  sign.  In  other 
words,  they  will  be  spaced  360  electrical  degrees  or  2ir  electrical 
radians  apart.  A  movement  of  the  armature  of  a  multipolar 
alternator  through  2ir  electrical  radians  is  equivalent,  so  far  as 
the  generation  of  voltage  is  concerned,  to  a  movement  of  the 
armature  of  a  two-pole  alternator  through  2ir  actual  or  space 
radians.  The  angular  velocity  of  the  armature  in  electrical 
radians  is  always  equal  to  co  =  2irf  where  /  is  the  frequency  of  the 
voltage  generated.  For  a  two-pole  machine,  frequency  is  equal 
to  the  revolutions  made  by  its  armature  per  second.  For  a 


ALTERNATING  CURRENTS  33 

multipolar  alternator  it  is  equal  to  revolutions  per  second  multi- 
plied by  pairs  of  poles. 

/  =  (revolutions  per  second)  X  (number  of  pairs  of  poles)     (7) 

Since  the  elements  of  the  armature  winding  for  a  multipolar 
alternator  are  2ir  electrical  radians  apart,  they  will  lie,  at  each 
instant,  in  the  same  relative  position  with  respect  to  north  and 
south  poles.  The  voltages  generated  in  them  therefore  will  be  in 
phase.  They  will  also  be  equal,  since  each  element  of  the  winding 
will  contain  the  same  number  of  turns.  Since  the  voltages  are 
both  equal  and  in  phase,  the  elements,  may  be  connected  either  in 
series  or  in  parallel.  Series  connection,  however,  is  the  more 
common  as  high  voltage  is  usually  desired.  The  standard  volt- 
ages most  commonly  used  for  alternators  are  2300  volts  and 
6600  volts.  The  frequencies  most  commonly  used  are  60  cycles 
and  25  cycles. 

The  armature  windings  of  alternators,  as  actually  built,  are 
not  placed  in  a  single  pair  of  slots  per  pair  of  poles,  as  shown  in 
Fig.  12,  but  are  distributed  among  a  number  of  pairs  of  slots  for 
each  pair  of  poles  and  are  thus  made  to  cover  a  considerable 
portion  of  the  armature  surface.  Spreading  out  the  winding  in 
this  way  increases  the  output  obtainable  from  an  armature  of 
given  size  and  also  improves  the  wave  form.  The  armature 
winding  of  a  single-phase  alternator  covers  about  two-thirds  of 
the  armature  periphery.  The  armature  windings  of  a  polyphase 
alternator  cover  the  entire  armature  surface,  but  the  winding 
for  any  one  phase  covers  only  a  portion  of  it. 

Commercial  alternators,  except  in  very  small  sizes,  are  uni- 
versally built  with  stationary  armatures  and  revolving  fields. 
(See  Fig.  70,  page  247.)  With  such  construction,  the  more 
complicated  part,  i.e.,  the  armature  winding,  is  stationary  and 
thus  is  relieved  from  all  stresses  except  those  caused  by  the 
current.  Moreover,  no  sliding  contacts  are  required  to  collect 
the  high-voltage  armature  current.  The  only  necessary  sliding 
contacts  are  those  for  the  field  excitation.  These  carry  a  com- 
paratively small  current  at  low  voltage. 

Strength  of  Current. — The  strength  of  an  alternating  current 
is  defined  in  terms  of  its  heating  effect.  An  alternating  current 
and  a  steady  current  are  said  to  be  equal  if  they  produce  heat 


34 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


at  equal  rates  when  they  flow  through  equal  resistances.  This 
assumes  that  the  distribution  of  current  over  the  cross  section 
of  the  conductor  is  the  same  in  both  cases. 

Let  i  be  the  instantaneous  value  of  the  alternating  current, 
i.e.,  its  strength  at  any  instant  of  time  t,  and  let  T  denote  the 
duration  of  a  complete  cycle.  The  rate  of  heating  in  a  given 
constant  resistance,  r,  at  any  instant  is  i2r  and  the  average 

rate  of  heating  during  a  cycle 

is 

1   rT  r   CT 

f  I    i2rdt  =  y  I    i2dt        (8) 

A  current  wave  (curve  a) 
is  plotted  in  Fig.  13.  Curve 
6  on  this  figure  is  obtained  by 
plotting  the  squared  ordinates 
of  curve  a.  The  integral 


FIG. 


jf- 


i2dt  is  the  mean  ordinate 
the  resistance  of  the  circuit, 
the  mean 


of  the  curve  b  multiplied   by  r, 

1    CT 

The  expression  -^  \    i2dt  is  the  mean  value  of  i2,  i. 
1  Jo 

square  of  the  instantaneous  values  of  the  current.  Denoting 
this  by  (mean  i2),  the  average  rate  of  heating  produced  by  a 
periodic  current  is  (mean  i2)  X  r.  For  a  steady  current,  /, 
the  rate  of  heating  as  given  by  Joule's  law  is/2r.  Hence,  in  order 
that  the  heating  effect  of  an  alternating  and  of  a  steady  current 
shall  be  equal  for  equal  resistances  (mean  i2)  must  equal  I2  or 
\/(niean  i2)  =  I.  From  this  it  follows  directly  that  a  periodic 
and  a  steady  current  are  equal  when  \/(mean  i2)  =  I  and  accord- 
ing to  this  definition,  the  rate  of  heating  by  a  periodic  current 
as  well  as  by  a  steady  or  direct  current  follows  Joule's  law. 

Ampere  Value  of  an  Alternating  Current. — An  alternating 
current  is  said  to  have  a  strength  of  one  ampere,  when  the  average 
rate  at  which  it  produces  heat  in  a  given  resistance  is  equal  to  the 
rate  at  which  heat  is  produced  by  a  steady  or  direct  current  of 
one  ampere  when  passing  through  an  equal  resistance.  Ob- 
viously according  to  this  definition,  the  ampere  value  of  an  alter- 
nating current  is  given  by 


ALTERNATING  CURRENTS  35 


tecifrf        (9) 

Volt  Value  of  an  Alternating  Voltage. — The  corresponding 
definition  of  a  volt  is  that  potential  difference  which  will  maintain 
one  ampere  alternating,  as  just  defined,  through  a  non-inductive 
resistance  of  one  ohm.  Obviously  an  alternating  voltage,  as  well 
as  an  alternating  current,  is  measured  by  the  square  root  of  its 
average  square  value. 

Advantages  of  Defining  the  Strength  of  Alternating  Currents 
by  the  Square  Root  of  their  Average  Square  Value. — The 
advantages  of  this  method  of  defining  the  strength  of  alternating 
currents  are: 

(a)  It  makes  it  possible  to  measure  any  alternating  current, 
of  whatever  wave  form,  by  its  heating  effect. 

(6)  It  makes  the  electrodynamometer  available  for  the  measure- 
ment of  all  alternating  currents  and  also  for  power  measurements. 

(c)  It  brings  alternating  currents  under  the  same  laws  of 
heating  as  steady  or  direct  currents. 

(d)  It  avoids  the  necessity  of  a  factor  in  the  expression  for 
power  which  would  be  different  for  each  wave  form. 

The  square  root  of  the  mean  square  of  the  instantaneous  values 
of  an  alternating  current  or  its  r^t-meaii-square  ialue,  usually 
abbreviated  into  r.m.s.,  is  known  as  its  effective ^value,  although  it 
is  sometimes  called  the  virtual  value. 

That  this  method  of  measuring  an  alternating  current  gives 
a  different  result  from  that  which  would  be  obtained  by  defining 
the  value  of  the  current  as  the  average  of  its  instantaneous  values 
is  obvious.  The  average  over  a  complete  cycle,  for  all  sym- 
metrical waves,  would  be  zero.  For  this  reason,  when  the 
average  value  of  an  alternating  current  is  mentioned,  the  average 
value  over  half  a  cycle  is  always  understood. 

The   average   value   of   an   alternating   current   is   given   by 

T 

Ml  (10) 


The  root-mean-square  value  is  given  by 

(11) 


36  PRINCIPLES  OF  ALTERNATING  CURRENTS 

These  are  equal  only  when  i  is  constant,  i.e.,  for  steady  cur- 
rents. The  average  value  of  the  current  is  equal  to  the  area 
enclosed  by  either  loop  of  the  curve  a,  Fig.  13,  page  34,  divided 
by  the  base  of  the  loop.  The  root-mean-square  value  of  the 
current  is  the  square  root  of  the  ratio  of  the  area  under  the  curve 
b  to  its  base. 

Relation  between  the  Root-mean-square  and  Average  Values 
for  Simple  Harmonic  or  Sinusoidal  Currents.  —  The  relation  be- 
tween the  root-mean-square  and  the  average  values  of  a  simple 
harmonic  current  is  easily  determined. 


~      =  0.7077W  (12) 


rmam  —  dt 

=  -7M=0.6377m  (13) 


Hence 

Iav.  =  0.637 

Ir.m.s.  ~    0.707 

Ir.m.s.  =    0.707 

lav.  0.637 


=  0.901  (14) 

=  1.11  (15) 


This  relation  between  the  average  and  root-mean-square  values 
of  an  alternating  current  may  be  made  clearer  by  the  state- 
ment that  if  a  simple  harmonic  current  of  one  ampere  were  flowing 
in  a  circuit,  the  average  of  its  instantaneous  values  would  be 
only  0.901  ampere  steady  current.  The  number  of  coulombs  in 
time  t  would  be  0.901*,  or  in  general  for  7  amperes  0.90172  cou- 
lombs. If  a  simple  harmonic  current  could  be  commutated  at 

T 

each  -_  second,  i.e.,  every  half  cycle,  the  weight  of  copper  de- 
posited by  it  per  second  would  be  only  0.901  of  that  deposited 
by  a  steady  current  of  the  same  number  of  amperes. 


ALTERNATING  CURRENTS  37 

Form  Factor. — The  ratio  of  the  root-mean-square  or  effective 
value  of  an  alternating  current  or  voltage  to  its  average  value  is 
its  form  factor.  (See  page  106.)  For  a  simple  harmonic  wave 
this  is  1.11. 

Measurement  of  the  Effective  or  Root-mean-square  Value  of 
a  Current  or  Voltage. — The  effective  or  root-mean-square  value 
of  an  alternating  current  may  be  measured  by  an  electrodyna- 
mometer.  Such  an  instrument,  in  its  simplest  form,  consists  of 
two  concentric  coils,  one  slightly  smaller  than  the  other.  The 
smaller  coil  is  mounted  inside  of  the  larger  in  such  a  manner  that 
it  may  turn  about  a  diameter  which  coincides  with  a  diameter  of 
the  larger  coil.  The  movement  of  the  smaller  coil  is  resisted  by 
two  spiral  springs  attached  to  its  shaft.  These  two  springs  also 
serve  to  connect  the  coil  in  circuit.  When  no  current  is  passing 
through  the  two  coils  the  springs  hold  the  plane  of  the  coils  at 
an  angle  of  about  45  degrees  with  each  other. 

If  the  two  coils  are  connected  in  series  and  a  current  is  passed 
through  them,  a  torque  will  be  developed  between  the  coils  which 
will  tend  to  make  the  movable  coil  turn  to  enclose  the  maximum 
flux.  For  any  relative  position  of  the  coils  this  torque  will  be 
proportional  to  the  product  of  their  ampere-turns.  Since  they 
are  in  series  and  carry  the  same  current,  the  torque  will  also  be 
proportional  to  the  square  of  the  current.  The  moving  coil  will 
deflect  until  the  torque  exerted  on  it  by  the  current  is  just 
balanced  by  the  opposing  torque  of  the  control  springs.  The 
direction  in  which  this  coil  moves  will  depend  upon  the  relative 
direction  of  the  current  in  the  coils.  The  relative  direction  of 
the  current  should  be  such  as  to  make  the  moving  coil  swing  so 
as  to  increase  the  angle  it  makes  with  the  other  coil.  This  will 
give  the  maximum  scale  range.  The  greatest  sensitivity  will 
occur  when  the  two  coils  are  at  right  angles.  If  the  current 
reverses  the  torque  exerted  between  the  coils  will  not  reverse 
since  the  relative  direction  of  the  currents  in  the  two  coils  will 
not  change. 

When  such  an  instrument  is  connected  in  an  alternating- 
current  circuit,  the  torque  exerted  between  the  coils  will  be  pro- 
portional, at  every  instant,  to  the  square  of  the  instantaneous 
current.  The  average  torque  will  be  proportional  to  the  average 
square  of  the  instantaneous  current  or  to  the  square  of  the 


38  PRINCIPLES  OF  ALTERNATING  CURRENTS 

effective  or  root-mean-square  current.  By  attaching  a  pointer 
to  the  moving  coil  and  providing  the  instrument  with  a  suitably 
graduated  scale,  it  may  be  made  to  indicate  effective  current. 
Since  for  any  given  position  of  the  coils,  the  torque  exerted  be- 
tween them  varies  as  the  square  of  the  current,  the  scale  cannot 
be  uniform.  In  general  it  will  be  more  or  less  cramped  at  the 
ends  and  open  in  the  middle. 

Many  refinements  are  made  in  the  instrument  as  actually 
constructed  to  increase  its  sensitivity  and  to  make  its  scale 
more  uniform. 

Since  the  control  springs  must  serve  to  carry  the  current  to 
and  from  the  moving  coil,  the  use  of  the  electrodynamometer 
type  of  ammeter  is  limited  to  the  measurement  of  very  small 
currents,  a  tenth  of  an  ampere  or  less,  unless  the  moving  coil 
is  shunted  with  an  inductive  shunt.  Inductive  shunts  are  some- 
times used  with  instruments  for  measuring  currents  up  to  five 
or  ten  amperes.  For  measuring  larger  currents,  the  moving 
coil  is  replaced  by  a  soft  iron  vane.  The  torque  acting 
to  deflect  this  vane  is  proportional  to  its  magnetic  moment 
and  to  that  component  of  the  field  produced  by  the  fixed 
coil  which  is  perpendicular  to  the  axis  of  the  vane.  If  the  vane 
is  operated  at  low  flux  density  and  is  made  of  suitable  iron,  its 
magnetic  moment  will  be  proportional  to  the  component  of  the 
field  which  lies  along  its  axis.  It  is  obvious  that  for  any  fixed 
position  of  the  vane,  the  torque  acting  to  deflect  it  will  be  pro- 
portional to  the  square  of  the  current  in  the  fixed  or  field  coil. 
If  the  current  reverses,  both  the  field  and  the  magnetic  moment 
of  the  vane  reverse.  The  direction  of  the  torque,  therefore, 
remains  unchanged.  When  an  instrument  arranged  in  this  way 
is  connected  in  series  with  an  alternating-current  circuit,  the 
vane  will  take  up  a  position  which  is  determined  by  the  average 
square  value  of  the  current.  The  limit  of  deflection  is  reached 
when  the  vane  is  perpendicular  to  the  plane  of  the  coil.  To 
extend  the  range  of  deflection,  the  coil  of  the  iron-vane  type  of 
instrument  is  inclined  forty-five  degrees  with  the  axis  about 
which  the  vane  turns. 

If  either  of  the  two  types  of  ammeter  is  wound  with  many 
turns  to  measure  small  currents  it  may  be  used  also  as  a  volt- 
meter. In  this  case  a  suitable  non-inductive  resistance  is  con- 


ALTERNATING  CURRENTS 


39 


nected  in  series  with  it  before  placing  the  instrument  across  the 
terminals  of  the  circuit  whose  potential  is  to  be  measured.  Since 
the  current  in  the  instrument  will  then  be  proportional  to  the 
voltage  at  every  instant,  the  instrument  may,  by  suitable 
calibration,  be  made  to  indicate  effective  or  root-mean-square 
voltage. 

All  alternating-current  ammeters,  except  those  for  measuring 
small  currents,  are  of  the  iron-vane  type.  The  best  volt- 
meters are  of  the  electrodynamometer  type.  The  electro- 
dynamometer  type  of  instrument  is  much  superior  to  instruments 
having  iron  vanes  but  its  use  is  limited  to  the  measurement  of 
small  currents. 

Vector  Representation  of  Simple  Harmonic  Currents  and 
Voltages. — The  simple  harmonic  current 

i  =  Im  sin  (co£  -+-  0) 

is  represented  graphically  by  the  curve  of  sines  plotted  in  Fig. 
14.  In  this  figure,  angles  are  plotted  as  abscissas  and  currents 
Y 


FIG.  14. 

as  ordinates.  The  abscissas  of  the  curve  are  also  marked  with 
the  corresponding  values  of  time  t.  The  ordinates  of  this  curve 
may  be  obtained  by  projecting  a  vector  OA,  revolving  in  a  posi- 
tive direction,  on  the  fixed  reference  line  0  Y.  The  vector  OA  must 
be  equal  in  length  to  the  maximum  value  of  the  current  and  it 
must  revolve  at  a  constant  angular  velocity  of  co  =  2irf  radians  per 
second.  Angles  are  reckoned  from  the  horizontal  reference  axis. 
The  counter-clockwise  direction  of  rotation,  as  was  stated  in 
Chapter  I,  is  always  considered  positive  for  rotating  vectors. 
Positive  angles,  therefore,  are  measured  in  a  counter-clockwise 
direction  and  are  angles  of  lead.  Negative  angles  or  angles  of 
lag  are  measured  in  a  clockwise  direction.  When  there  are 
several  vectors  it  is  sometimes  convenient  to  fix  the  vectors  and 


40 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


rotate  the  axes  of  reference.  When  this  is  done  a  clockwise 
rotation  of  the  axes  is  positive. 

The  vector  OA,  Fig.  14,  makes  one  revolution  in  a  time  T  =  , 

second,  the  periodic  time  of  the  current,  and  occupies  a  position 
perpendicular  to  the  axis  OF,  the  axis  on  which  the  projections 
are  taken,  at  a  time  determined  by  the  condition  (ut  +  6)  =*  0. 
Since  the  instantaneous  values  of  the  current  are  determined 
from  the  projections  of  the  vector  OA  on  the  perpendicular  axis 
OF,  this  vector  OA  may  be  considered  to  represent  the  current 
i  =  Im  sin  (a>t  +  0).  The  phase  of  the  current  is  so  determined 
that  at  a  time  t  =  0  the  vector  OA  =  Im  makes  a  positive  angle 
0,  its  phase  angle,  with  the  horizontal  reference  axis  from  which 
both  angles  and  time  are  reckoned.  The  rotating  vector  exactly 
represents  the  current  in  both  magnitude  and  phase,  since  its 
projection  on  the  vertical  axis  OF  at  any  instant  represents 
the  magnitude  of  the  current  at  that  instant  and  the  angle 
(at  +  6)  which  it  makes  with  the  horizontal  axis  represents  the 
phase  of  the  current  at  that  instant. 


FIG.  15. 

When  considering  several  simple  harmonic  currents  and  volt- 
ages of  the  same  frequency,  each  may  be  represented  in  both 
magnitude  and  phase  by  a  revolving  vector  of  proper  length 
and  position,  all  vectors  being  drawn  from  the  same  point. 

Consider  a  circuit  consisting  of  two  branches  in  parallel 
carrying  simple  harmonic  currents  i\  and  i?  given  by.  the  following 
equations : 

i!  =  Imi  sin  M  +  0i)  (16) 

iz  =  Imi  sin  (<d  +  62)  (17) 

These  two  currents  and  their  resultant  *0  are  plotted  in  Fig.  15. 


ALTERNATING   CURRENTS  41 

The  resultant  current,  at  any  instant,  is  equal  to  the  algebraic 
sum  of  the  instantaneous  values  of  the  two  component  currents 
i\  and  i2.  That  is, 

i<>  =  i\  +  it 

Referring  to  Fig.  15,  the  resultant  current  xc  at  the  instant 
of  time  marked  tf  is  equal  to  the  algebraic  sum  of  the  instanta- 
neous values  xa  and  xb  of  the  two  component  currents.  The 
resultant  current  at  the  time  t1  is  equal  to  the  projection  of  7m0, 
the  resultant  of  the  two  component  vectors  Im\  and  Im2,  on  the 
vertical  axis  at  time  if.  In  the  left-hand  part  of  the  figure  the 
three  revolving  vectors  are  shown  in  the  positions  they  occupy 
when  t  =  0. 

At  the  instant  of  time  t 

io  =  Imo  sin  (at  -f  00) 

=  Iml  sin  (ut  +  Oi)  +  /„,  sin  (ut  +  02)        (18) 

The  angle  (0i  —  62)  is  the  difference  in  phase  between  the  two 
component  currents  whose  maximum  values  are  Im\  and  7m2. 
Im2  lags  Imi  by  an  angle  (0i  —  02),  i.e.,  it  goes  through  its  maxi- 

T 

mum  positive  value   (0i  —  02)  radians    or  (0i  —  02)  ^-  seconds 

later  than  Im\.  The  phase  differences  between  the  components 
Im\  and  7m2  and  the  resultant,  whose  maximum  value  is  7m0,  are 
respectively  (0i  —  00)  and  (02  —  00).  The  resultant  goes  through 
its  maximum  value  (0i  —  00)  radians  behind  Imi  and  (00  —  62) 
radians  ahead  of  7m2.  (0i  —  00)  is  its  angle  of  lag  with  respect 
to  Imi  and  (00  —  02)  is  its  angle  of  lead  with  respect  to  7m2. 
Expanding  the  sine  terms  in  equation  (18)  gives 

io  =  7m0  (sin  o>£  cos  00  +  cos  ut  sin  00) 
=  Imi  (sin  wt  cos  0i  +  cos  ut  sin  0i) 

+  7m2  (sin  co£  cos  02  +  cos  at  sin  02)     (19) 

Equation  (19)  must  hold  for  all  values  of  t. 

When  t  =  0,  sin  <at  =  0  and  cos  ait  =  1.     In  this  case 

io  =  Imo  sin  00  =  Im\  sin  0i  +  7m2  sin  02  (20) 

T 
When  t  =  -T-,  sin  <*>t  =  1  and  cos  erf  =  0.     In  this  case 

iQ  =  Im0  cos  00  =  Imi  cos  0i  +  7m2  cos  02  (21) 


42  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Equations  (20)  and  (21)  represent  two  right-angle  components 
of  the  resultant  vector  7m0.     Its  magnitude  must  therefore  be 


/mo  =       C/mi  cos  0i  +  I  mz  cos  02)2  +  (/mi  sin  0i  +  7m2sin  02)2 
=  V(Sf  7m  cos  0)2  +  (SJ  7TO  sin  0)2  (22) 

mi  sin  0i  +  7m2  sin  02 


=     ml  cos    !          m2  2 


S}  7TO  cos  0 

Equations  (22)  and  (23)  are  not  limited  to  two  currents  or  in 
general  to  two  revolving  vectors.  They  may  be  extended  to 
apply  to  any  number.  If  there  are  k  currents  of  the  same  fre- 
quency to  be  added 


7MO  =       (S*!/«  cos  0)2  +  (S{/»  sin  0)2  (24) 

2\Im  sin  0 

tan  "«  = 


Equations  (22)  and  (23)  and  also  equations  (24)  and  (25)  may 
be  obtained  more  easily  by  the  use  of  the  vector  or  complex 
expressions  for  the  currents.  Take  the  axis  of  time  as  the 
axis  of  reals  (See  page  6  Chap.  I)  and  consider  the  vectors  at 
the  instant  t  =  0.  At  this  instant  7w0,  Im\  and  7w2  make  angles 
00,  0i  and  02  with  the  axis  of  reals.  Their  complex  expressions  at 
this  instant  are  therefore 

ImO    =   ImO  (COS   00   +  j  SU1   00)    =    tt0   +  jb0 

I  mi  =  I  mi  (cos  0!  +  j  sin  0i)  =  ai  +  jbi 
7m2  =  Im2  (cos  02  -f-  j  sin  02)  =  a2  +  jbz 

7TO0    =    Iml   ~\~  Im2 

aQ  +  jb0  =  (ai  +  j  61)  +  (o2  +  jbz) 


+  a2)2  +  (61 
tan  00  =  - 


ALTERNATING  CURRENTS  43 

If  there  are  k  vectors 


tan  00  = 


a2         .     . 

+  &2  +     .    .     + 


+  «2  +     -    •     +  a* 
Jmo  =    2*  /„,  cos  0  -h  j  2*  JTO  sin 


/«o  =       (Si/m  cos  0)2  +  (S*/w  sin  0)* 

2*  Im  sin  0 

tan  t/o  =  ^irr  ---  ;r 

2*  Im  cos  0 

Since  the  relations  existing  among  the  maximum  values  of 
rotating  vectors  and  also  among  their  phase  angles  are  inde- 
pendent of  the  instant  of  time  considered,  it  is  not  necessary 
to  take  t  =  0  when  adding  currents  or  voltages.  The  value  of  t 
which  will  give  the  simplest  and  easiest  solution  should  be  chosen. 
Usually  this  is  t  =  0,  although  it  is  better,  in  many  cases,  to  give 
t  such  a  value  that  one  of  the  vectors  will  lie  along  the  axis  of 
reals.  When  the  solution  of  a  problem  involves  both  currents 
and  voltages  and  there  is  a  common  vector,  such  as  current  in  a 
series  circuit  or  voltage  in  a  parallel  circuit,  this  common  vector 
may  well  be  taken  for  the  axis  of  reals.  For  example,  in  a  circuit 
consisting  of  a  number  of  branches  in  parallel  the  same  voltage  is 
impressed  across  each.  It  is  common  to  the  currents  in  the 
branches.  In  general  it  should  be  taken  as  the  axis  of  reals.  In 
the  case  of  a  series  circuit  the  current  is  the  same  in  all  parts. 
In  general,  for  a  series  circuit,  the  current  should  be  taken  for  the 
axis  of  reals. 

In  most  problems  arising  in  alternating-current  work,  interest 
centers  on  the  effective  values  of  currents  and  voltages,  i.e.,  root- 
mean-square  values.  Maximum  values  are  seldom  used  or 
desired.  Since  root-mean-square  values  and  maximum  values 
are  proportional  (Jro  =  \/2  /r.m.«.)  it  is  customary  to  write  the 
vector  expressions  for  all  alternating  currents  and  voltages  in 

trms  of  their  root-mean-square  or  effective  values. 
An  Example  of  Addition  of  Currents.  —  A  load  on  a  certain  line 
consists  of  an  induction  motor  in  parallel  with  a  synchronous 
motor.  Since  the  motors  are  in  parallel,  the  voltage  across  their 
terminals  must  be  equal  at  every  instant.  The  vectors  repre- 
senting the  voltages  impressed  across  the  terminals  of  the  motors 
must  therefore  be  in  phase  and  must  also  be  equal  in  magnitude. 


44  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  induction  motor  takes  a  current  of  100  amperes  (r.m.s.) 
which  lags  the  impressed  voltage  by  30  degrees.  The  current 
taken  by  the  synchronous  motor  is  50  amperes.  This  leads 
the  impressed  voltage  by  60  degrees.  What  is  the  resultant 
current  which  must  be  supplied  to  these  motors  in  parallel  and 
what  is  its  phase  with  respect  to  the  voltage  impressed  across 
their  terminals? 

Take  the  voltage  as  the  axis  of  reals.  The  vector  expressions 
for  the  voltage  and  currents,  using  root-mean-square  values,  will 
then  be 

V  =  7(1  +  JO) 

/»•  =  100  (cos  30°  -  j  sin  30°) 
=  86.6  -  j'50 

/.  =  50 (cos  6,0°  +  j  sin  60°) 
=  25.0  +  J43.3 

/O    =    Ii   +  Is 

=  (86.6  -  j'50),+  (25.0  +  J43.3) 
=  111.6  -J6.7 


70  =  V(H1.6)2  +  (6.7)2 
=  111.9  amperes  r.m.s. 

tan  00=  -yfre  =  -°-0600 
80  =  —3.44  degrees 

The  resultant  current  70  is  111.9  amperes  and  it  lags  the  voltage 
impressed  across  the  motors  by  3.44  degrees.     In  other  words, 

it  goes  through  its  cycle  3.44  degrees  or  3.44  X  second 

ouu  /\  J 

later  than  the  voltage,  /  being  the  frequency  of  the  circuit. 
Another  Example. — A  two-pole,  three-phase,  sixty-cycle  alter- 
nator has  three  exactly  similar  windings  on  its  armature,  spaced 
120  degrees  from  one  another  and  each  containing  100  turns. 
Two  of  these  windings  are  connected  in  series  with  each  other  in 
such  a  way  that  the  voltage  between  their  free  terminals  is  equal 
to  the  difference  between  their  induced  voltages.  If  these  two 
windings  in  series  are  connected  to  a  constant  non-inductive 
resistance  of  10  ohms,  what  are  the  maximum  and  effective  values 
of  the  current  in  the  resistance?  How  much  power  is  absorbed 


ALTERNATING  CURRENTS  45 

by  the  resistance?  What  is  the  phase  relation  of  the  current  in 
the  resistance  to  the  voltage  impressed  across  its  terminals? 
What  is  the  phase  relation  between  the  voltage  across  the  two 
windings  in  series  and  the  voltage  induced  in  each  winding? 
The  field  flux  of  the  alternator  per  pole  is  106  maxwells.  This  is 
distributed  in  such  a  way  that  the  flux  linking  each  armature 
winding  varies  sinusoidally  with  time. 

The  voltage  induced  in  any  armature  winding  is 


where  #  is  the  flux  linking  the  winding  at  the  time  t  and  N  is  the 
number  of  turns  in  the  winding.  According  to  the  assumption 
that  the  flux  linking  each  armature  winding  varies  sinusoidally 
with  time, 

<P  =  <pm  sin  cot 

where  co  is  the  angular  velocity  of  the  armature  and  is  equal  to 
2-n-  times  the  frequency  of  the  voltage  induced  in  it.  For  a  two- 
pole  alternator,  the  frequency,  /,  is  equal  to  the  speed  of  its 
armature  in  revolutions  per  second. 


m  sin  at 

=     —Ci)N<f>m    COS   tot 

The  maximum  value  of  this  voltage  is  obviously 
The  effective  or  root-mean-square  value  is 


\/2 

=  4.44  Nfom  abvolts 
=  4.44  NfamlQ-*  volts 

The  voltage  induced  in  each  winding  of  the  alternator  is 
therefore 

E=  4.44  X  100  X  60  X  106  X  10~8 
=  266.4  volts  effective. 

Let  the  armature  windings  be  numbered  1,  2,  3  and  assume  that 
the  direction  of  rotation  of  the  armature  is  such  that  the  phase 


46  PRINCIPLES  OF  ALTERNATING  CURRENTS 

order  is  also  1,  2,  3,  i.e.,  the  voltages  induced  in  the  armature 
winding  go  through  their  cycles  in  the  order  1,  2,  3.  The  three 
voltages  must  be  120  degrees  apart  in  time  phase  since  the  arma- 
ture windings  are  spaced  120  degrees  apart  on  the  armature. 
Let  the  windings  1  and  2  be  the  ones  connected  to  the  resistance. 
Take  the  voltage  induced  in  phase  1  as  along  the  axis  of  reals. 
The  vector  expressions  for  the  three  voltages  will  then  be 

li  =  #i(cos  0°  -  j  sin  0°) 

=  266.4(1  -  JO) 

=  266.4  -  JO 
Ez  =  #2(cos  120°  -  j  sin  120°) 

=  266.4(-0.5  -j  0.866) 

=  -133.2  -  J230.6 
E3  =  E3(cos  240°  -  j  sin  240°) 

=  266.4  ( -0.5  +j  0.866) 

=  -133.2  +j  230.6 

The  voltage  between  the  terminals  of  windings  1  and  2, 
connected  in  such  a  way  that  their  voltages  subtract,  is 

E0=Ei-E2  =  (266.4  -  jQ)  -  (-133.2 -.7  230.6) 

=  399.6  +  j  230.6 
E0=  V(399.6)2   +  (230.6)2 
=  461.4  volts. 

tan  «° = « =  °-578 

00  =  30  degrees. 

The  voltage  E0  =  E^  —  E2  is  therefore  equal  to  461.4  volts 
and  leads  the  voltage  EI  by  30  degrees.  It  leads  the  voltage  E% 
by  120  +  30  =  150  degrees.  The  expression  for  its  instantane- 
ous value  is 

eo  =  ^/2  X  461.4  sin  (27r60<  +  30°) 
=  652.4  sin  (377*  +  30°) 

Since  the  circuit  contains  nothing  but  resistance,  the  voltage 
impressed  across  it  must  be  equal  to  the  resistance  drop  at  each 
instant.  Since  the  resistance  is  constant,  the  current  i  through 


ALTERNATING  CURRENTS  47 

the  resistance  must  be  proportional  to  the  voltage  at  each  in- 
stant.    Hence 

E0  sin  (at  +  30°) 


i  = 


J-r.m.s. 


r 
Er,m          461.4 


r  10 

=  46. 14  amperes. 

Im  =  \/2  X  46.14. 
=  65.24  amperes. 

Since  the  current  is  in  phase  with  the  voltage,  the  expression 
for  the  instantaneous  current  is 

i  =  V2  X  46.14  sin  (377  t  +  30°  ±  0°) 
=  65.24  sin  (377*  +  30°) 

The  complex  expression  for  the  current  in  terms  of  its  root- 
mean-square  value  is 

_  _  #0  _  399.6  +  J230.6 

:T:  10 

=  39.96  +  J23.06 

The  vector  expression  for  the  current  may  also  be  found  by 
making  use  of  the  operator  which  will  rotate  a  vector  through  a 
given  angle.  Since  the  current  is  in  phase  with  the  voltage,  it 
must  be  displaced  from  the  axis  of  reals  by  the  same  angle  as  the 
voltage  EQ  producing  it,  i.e.,  by  an  angle  of  30  degrees.  The 
operator  which  will  displace  a  vector  to  which  it  is  applied  by  an 
angle  0  is  (cos  0  +  j  sin  6).  (See  Chapter  I,  page  7.)  There- 
fore, 

7  =  7  (cos  6  +  j  sin  B) 

=  46. 14  (cos  30°  +  jsin30°) 

=  46.14  (0.866  +  jO.500) 

=  39.96  +  j  23.07 

From  the  definition  of  the  effective  or  root-mean-square  value 
of  a  current  it  follows  that  the  average  power  absorbed  in  the 
resistance  is 

P  =  Pr 

=  (46. 14)2  X  10 
=  21,290  watts. 


CHAPTER  III 

POWER  WHEN  CURRENT  AND  VOLTAGE  ARE  SINUSOIDAL 

Power  Absorbed  and  Delivered  by  a  Circuit. — The  problems 
concerning  power  which  arise  in  dealing  with  direct-current 
circuits  are  usually  simple  and  there  is  seldom  any  doubt  as  to 
whether  power  is  absorbed  or  delivered.  The  conditions  are  not 
always  so  simple  with  alternating  currents.  It  is  necessary 
therefore  to  determine  definitely  the  conditions  under  which 
power  is  absorbed  and  delivered.  The  fundamental  conditions 
are  the  same  for  direct-  and  alternating-current  circuits. 

When  a  direct-current  dynamo  acts  as  a  generator,  the  current 
flow  is  from  the  plus  to  the  minus  terminal  through  the  external 
circuit  and  it  is  from  the  minus  to  the  plus  terminal  through  the 
armature  of  the  dynamo.  The  machine  is  delivering  power. 
The  load  is  absorbing  power.  Through  the  load  the  current  flow 
is  from  a  higher  to  a  lower  potential  or  in  the  direction  of  decreas- 
ing potential,  i.e.,  it  is  in  the  direction  of  the  potential  drop. 
Through  the  dynamo  the  current  flow  is  from  a  lower  to  a  higher 
potential  or  in  the  direction  of  increasing  potential,  i.e.,  it  is  in 
the  direction  of  the  potential  rise.  When  the  dynamo  acts  as  a 
motor,  the  current  flows  through  the  armature  from  the  plus  to 
the  minus  terminal  or  in  the  direction  of  the  potential  drop.  In 
this  case  power  is  being  absorbed  by  the  dynamo.  In  general, 
if  there  is  a  rise  in  potential  in  the  direction  of  the  current 
flow,  power  is  being  generated  and  power  is  delivered.  If  there 
is  a  drop  or  fall  in  potential  in  the  direction  of  current  flow, 
power  is  absorbed. 

The  conditions  under  which  power  is  absorbed  or  power 
delivered  by  an  alternating-current  circuit  are  exactly  the  same 
as  those  for  a  direct-current  circuit.  Due,  however,  to  the  differ- 
ence in  phase  of  the  current  and  voltage  in  most  alternating-cur- 
rent circuits,  the  current  is  seldom  in  the  direction  of  either  the 
voltage  rise  or  the  voltage  fall  during  all  parts  of  a  cycle.  Con- 
sequently, power  is  absorbed  during  part  of  each  cycle  and  de- 

48 


POWER  49 

livered  during  the  remaining  part.  If  the  average  power 
absorbed  during  a  cycle  is  greater  than  the  average  power 
delivered,  the  net  effect  is  power  absorbed.  If  the  average 
power  delivered  during  each  cycle  is  greater  than  the  average 
power  absorbed,  the  net  effect  is  power  delivered. 

Instantaneous  Power. — The  instantaneous  power,  p,  in  a 
circuit  is  given  by  the  product  of  the  instantaneous  values  of  the 
current  and  voltage. 

p  =  ei 

If  e  is  a  voltage  rise  in  the  direction  of  current  flow,  p  represents 
power  delivered.  When  e  is  a  fall  in  voltage  in  the  direction 
of  current  flow,  power  is  absorbed.  Whether  power  delivered 
or  power  absorbed  is  considered  positive  or  negative  will  depend 
upon  the  convention  adopted  for  the  positive  direction  of  a 
voltage  rise. 

If  the  voltage  e  is  considered  positive  when  it  represents  an 
actual  rise  in  voltage  in  the  direction  assumed  positive  for  current 
flow,  the  expression  p  =  ei  represents  power  delivered.  When 
the  product  of  e  and  i  is  positive,  power  is  actually  delivered, 
since  e  will  then  actually  be  increasing  in  the  direction  of  current 
flow.  When  the  product  of  e  and  i  is  negative,  e  will  actually  be 
decreasing  in  the  direction  of  current  flow  and  power  will  be 
absorbed.  In  the  latter  case  there  is  a  fall  of  potential,  i.e. 
a  negative  rise  in  potential,  in  the  direction  of  current  flow  and 
the  power  delivered  is  negative.  Negative'  power  delivered  is 
power  absorbed.  /According  to  this  convention  a  voltage  rise 
in  the  positive  direction  of  current  flow  is  positive.  The  negative 
of  a  voltage  rise  is  a  voltage  fall.! 

If  the  positive  direction  for  a  voltage  rise  is  taken  opposite 
to  the  positive  direction  for  current  flow,  the  expression  p  =  ei 
will  represent  power  absorbed.  Power  will  actually  be  absorbed 
when  p  =  ei  is  positive  and  it  will  actually  be  delivered  when 
p  =  ei  is  negative.  According  to  this  convention  a  voltage 
drop,  or  fall  in  potential  in  the  positive  direction  of  current  flow, 
is  considered  positive. 

Curves  of  e,  i  and  p  =  ei  are  plotted  in  Fig.  16.  If  e  represents 
a  voltage  rise,  considered  positive  in  the  positive  direction  of 
current  flow,  power  will  be  delivered  when  e  and  i  have  the  same 
sign,  i.e.,  when  both  are  positive  or  both  are  negative.  There 

4 


50  PRINCIPLES  OF  ALTERNATING  CURRENTS 

will  then  be  an  actual  rise  in  voltage  in  the  direction  of  current 
flow.  When  e  and  i  have  opposite  signs  power  will  be  absorbed. 
In  this  case  there  will  be  an  actual  fall  in  voltage  in  the  direction 
of  current  flow.  Power  delivered  will  be  positive  and  power 
absorbed  will  be  negative. 


Dashed  Line  Voltage 
Full  Line  Current 
Dotted  Line  Power 


FIG.  16. 

When  only  power  delivered  is  to  be  considered,  it  is  best  to 
take  the  positive  direction  for  a  voltage  rise  in  the  direction 
assumed  positive  for  current  flow,  since  this  convention  makes 
power  delivered  positive  and  avoids  the  use  of  negative  signs 
before  the  expressions  for  power  delivered.  This  convention  is 
best  and  also  most  logical  when  power  delivered  and  power 
absorbed  are  involved  in  the  same  problem.  When,  however, 
only  power  absorbed  is  to  be  considered,  it  is  best  to  adopt  the 
convention  that  voltage  drops  are  positive  in  the  direction 
assumed  positive  for  current  flow,  since  then  power  absorbed  is 
positive  and  the  necessity  of  using  minus  signs  before  the  expres- 
sions for  power  absorbed  is  avoided. 

Average  Power.— The  power  delivered  or  absorbed  by  an 
alternating-current  circuit  is  the  average  power  considered  over 
a  complete  cycle,  i  It  is  seldom  that  the  instantaneous  power  is 
of  interest.  The  average  power  is  given  by 

1  CT 
P  =  T      eidt  (1) 


POWER  51 

This  expression  gives  the  mean  ordinate  of  the  power  curve,  i.e., 
the  dotted  curve  in  Fig.  16,  and  is  equal  to  the  net  area  enclosed 
by  the  power  curve  divided  by  its  base.  In  order  to  evaluate  the 
integral,  it  is  necessary  to  know  the  form  of  the  functions  deter- 
mining the  voltage,  e}  and  the  current,  i. 

Power  when  the  Voltage  and  Current  Waves  are  both  Sinu- 
soidal.— Assume  the  voltage  and  current  waves  to  be  sinusoidal. 

CASE  I.  VOLTAGE  AND  CURRENT  IN  PHASE. — Let 

e  =  Em  sin  co£ 
i  =  Im  sin  ut 

p  =  ei  =  Emlm  sin2  at  =EmIm  i     -~i*^       (2) 

1  CT  Emlm           1  CT  E  I 
Average  power  =  P  =  —  I        m      dt I        n'  m  cos  2ut  dt 

=  EmIm  jEn_   •     7^ 

2        ~  Vl    "  V? 
=  EI  (3) 

where  E  and  /  without  subscripts  are  the  root-mean-square  or 
effective  values  of  the  voltage  and  current.  Therefore,  when 
the  voltage  and  current  are  in  phase,  the  average  power  is  equal 
to  the  product  of  the  effective  values  of  voltage  and  current.  It 
is  equal  to  the  effective  volt-amperes,  i.e.,  the  product  of  the 
effective  voltage  and  effective  current. 

The  expression  for  the  instantaneous  power,  p,  (Equation  2), 
may  be  written 

p  =  p  _    _^»  cos  2wt  =  p  -  p  Cos  2wt  (4) 

This  equation  and  the  equations  for  voltage  and  current  are 
plotted  in  Fig.  17.  It  should  be  noted  that  the  power  curve  is 
sinusoidal  in  form  and  has  double  frequency  as  compared  with 
the  voltage  and  current.  Its  axis  of  symmetry  is  at  a  distance 
P,  equal  to  the  average  power,  above  the  axis  of  the  voltage  and 
current  curves.  Its  amplitude  is  equal  to  one-half  the  maximum 
volt-amperes  or  is  equal  to  the  effective  volt-amperes.  Effective 
volt-amperes  are  equal  to  average  power  when  the  voltage  and 
current  are  in  phase.  Although  the  power  is  not  constant,  its 


52 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


flow  is  always  in  the  same  direction  when  the  voltage  and  current 
are  in  phase.  The  corresponding  instantaneous  values  of  voltage 
and  current  are  always  of  the  same  sign.  The  current  flow  is 
therefore  always  either  in  the  direction  of  the  voltage  rise  or  in 
the  direction  of  the  voltage  drop,  according  as  e,  in  the  equation 
for  voltage,  represents  a  voltage  rise  or  a  voltage  drop.  If  e 


t=o 


FIG.  17. 

represents  a  voltage  rise,  the  expression  for  average  power  gives 
the  power  delivered.  If  e  represents  a  voltage  drop,  the  expres- 
sion for  average  power  gives  the  power  absorbed.  If  e,  represent- 
ing a  voltage  rise,  and  i  are  opposite  in  phase  the  expression  for 
power  will  be  negative  and  will  represent  power  absorbed.  If  e, 
representing  a  voltage  drop,  and  i  are  opposite  in  phase  the 
expression  for  power  represents  power  delivered. 

CASE  II.  VOLTAGE  AND  CURRENT  IN  QUADRATURE.  —  Let 

e  =  Em  sin  cot 
i  =  Im  sin  (at  -  90°) 

p  =  ei  =  Enlm(sm2  ut  cos  90°  -  sin  cat  cos  co£  sin  90°) 
=  Emlm(0  —  sin  co£  cos  wt) 


-  sin  2«0 


(5) 


1  CTF   1 
Average  power  =  P  =  ~  I      i2±2  (0  -  sin  2wt)dt 

•*•  t/o         ^ 
=  0 

When  the  voltage  and  current  are  in  quadrature  the  average 
power  is  zero. 

The  curves  for  instantaneous  power  (equation  (5))  and  for  the 


POWER 


53 


instantaneous  voltage  and  current  are  plotted  in  Fig.  18.  The 
power  curve  is  again  a  simple  harmonic  curve  of  double  frequency 
with  respect  to  the  voltage  and  current  and  it  still  has  an  ampli- 
tude equal  to  the  effective  volt-amperes,  but  its  axis  of  sym- 
metry now  coincides  with  the  axis  of  symmetry  of  the  voltage  and 
current.  This  latter  condition  follows  from  the  fact  that  the 
average  power,  P,  is  zero. 


t-o 


FIG.  18. 

Although  the  average  power  is  zero,  the  power  at  any  instant 
is  not  zero  except  at  four  points  during  each  cycle.  There  is  an 
oscillation  of  power  between  the  source  and  the  load,  the  average 
value  of  which  is  zero.  The  amplitude  of  this  oscillation  is  equal 
to  El,  the  effective  volt-amperes  of  the  circuit.  If  e  represents 
a  voltage  rise,  the  positive  power  loops  represent  power  delivered. 
The  negative  power  loops  represent  power  absorbed.  During 
two  quarters  of  each  cycle,  power  is  delivered  by  the  circuit. 
During  the  other  two  quarters  of  each  cycle  an  equal  amount  of 
power  is  absorbed.  The  power  which  is  absorbed  is  stored  as 
kinetic  energy  either  in  the  rotating  part  of  a  motor  or  generator 
or  in  a  magnetic  field  produced  by  the  current,  or  it  may  be  stored 
as  potential  energy  in  the  electrostatic  field  of  a  condenser. 
This  stored  kinetic  or  potential  energy  is  given  back  to  the  circuit 
as  dynamic  electrical  energy. 

CASE  III.  GENERAL  CASE.  VOLTAGE  AND  CURRENT  NEITHER 
IN  PHASE  NOR  IN  QUADRATURE. — Let 

e  =  Em  sin  ut 

i  =  Im  sin  (at  —  e) 


54 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


where  6  is  neither  0  nor  90  degrees. 

p  =  ei  =  Emlm  sin  ut  sin  (wt  —  6)  (6) 

Since  2  sin  a  sin  0  =  cos  (a  —  |8)  —  cos  (a  +  0),  equation  (6), 
may  be  written 


P  = 


EmL 


cos  (  +  (9)  -  cos  (2co£  - 


(7) 


i  rrjr  f 
Average  power  =  P  =  _  I    fl^fcos  (  +  0) 

TJo       2 


-C'  TO-* 


cos 


i      f\          -     w* 

+  0=  —  ^ 


—  cos (2ui  — 
L 


V2"  V2 


COS0 


cos  0 


(8) 


The  average  power  is  equal  to  the  product  of  the  effective 
values  of  voltage  and  current  multiplied  by  the  cosine  of  the 
phase  angle  between  the  voltage  and  current.  When  the  phase 
angle  is  zero,  P  =  EI  cos  6  becomes  EI.  When  the  phase  angle 
is  90  degrees,  P  =  EI  cos  6  becomes  zero. 

Equation  (7)  for  the  instantaneous  power  may  be  written 

p  =  EI  cos  6  -  EI  cos  (2co/  -  0) 

=  P  -  EI  cos  (2ut  -  6)  (9) 

From  equation  (9)  it  may  be  seen  that,  in  the  general  case  as 
well  as  in  the  two  special  cases  first  considered,  the  power  curve 

p 


t-o 


1!). 


is  a  double  frequency  curve.  As  in  the  first  two  cases  it  has  an 
amplitude  equal  to  the  effective  volt-amperes  and  its  axis  is  dis- 
placed from  the  axis  of  voltage  and  current  by  a  distance  equal  to 
the  average  power.  Curves  of  power,  voltage  and  current  are 
plotted  in  Fig.  19. 


POWER  55 

If  e  is  taken  as  a  voltage  rise,  equation  (8)  represents  power 
delivered.  Power  will  actually  be  delivered  when  El  cos  6 
is  positive.  Power  will  actually  be  absorbed  when  El  cos  0  is 
negative.  It  will  be  positive  when  0  is  less  than  90  degrees,  since 
the  cosine  of  an  angle  which  is  less  than  90  degrees  is  positive. 
It  will  be  negative  when  6  is  greater  than  90  degrees  (but  less  than 
270  degrees) ,  since  the  cosine  of  an  angle  which  is  greater  than  90 
degrees  (but  less  than  270  degrees)  is  negative.  Since  an  angle  of 
lag  of  6  degrees  is  the  same  as  an  angle  of  lead  of  360  minus  0 
degrees  and  similarly  an  angle  of  lead  of  6  degrees  is  the  same  as 
an  angle  of  lag  of  360  minus  0  degrees,  it  is  customary  to  use  the 
smaller  of  the  two  angles  when  expressing  the  phase  difference 
of  a  current  and  a  voltage.  Equal  angles  of  lead  and  lag  pro- 
duce the  same  effect  so  far  as  average  power  is  concerned. 

When  0  =  0  the  power  curve,  Fig.  19,  is  entirely  above  the 
axis  of  current  and  voltage.  This  corresponds  to  the  conditions 
shown  in  Fig.  17  page  52  for  Case  I  where  the  voltage  and  current 
are  in  phase.  When  0  is  greater  than  zero  and  less  than  90  de- 
grees, the  axis  of  the  power  curve  is  displaced  in  a  positive  direc- 
tion from  the  axis  of  the  voltage  and  current  curves.  In  this  case 
(assuming  E  represents  a  voltage  rise)  power  is  delivered.  If 
0  is  greater  than  90  degrees,  the  axis  of  the  power  curve  is  dis- 
placed in  a  negative  direction  from  the  axis  of  the  voltage  and 
current  curves.  In  this  case  the  average  power  is  negative  and 
power  is  absorbed.  If  0  is  equal  to  90  degrees,  power  is  neither 
delivered  nor  absorbed.  This  corresponds  to  Case  II. 

Volt-amperes ;  Apparent  or  Virtual  Power. — The  volt-amperes, 
or  apparent  or  virtual  power  of  a  circuit,  is  equal  to  the  product 
of  the  effective  or  root-mean-square  values  of  voltage  and  current. 
This  product  is  not  equal  to  the  true  power  except,  when  the 
voltage  and  current  are  in  phase.  Although  volt-amperes  do 
not  represent  true  power  Except  in  the  case  mentioned,  a  know- 
ledge of  the  volt-amperes  of  a  circuit  is  usually  of  considerable 
importance,  since  volt-amperes  and  not  watts  determine  the 
limit  of  output  of  most  electrical  apparatus.  The  limit  of  output 
of  all  alternating-current  apparatus,  such  as  motors,  generators, 
transformers,  etc.,  is  determined  chiefly  by  the  rise  in  temperature 
produced  in  the  windings.  The  increase  in  temperature  is 
caused  principally  by  the  core  and  copper  losses.  Core  losses 


56  PRINCIPLES  OF  ALTERNATING  CURRENTS 

depend  upon  frequency  and  flux  density  and  are  fixed  by  the 
operating  voltage  and  frequency.  Copper  losses  are  determined 
by  the  current.  For  fixed  voltage  and  current  these  losses  are 
practically  independent  of  the  phase  relation  between  current 
and  voltage.  The  limit  of  output  of  a  generator,  motor  or  trans- 
former, therefore,  is  determined  by  the  volt-amperes  which 
produce  the  limiting  rise  in  temperature  in  the  windings.  All 
types  of  alternating-current  apparatus,  which  can  operate  with 
different  phase  angles  between  current  and  voltage,  i.e.,  at 
different  power  factors,  are  rated  in  volt-amperes.  Full  load 
is  reached  when  they  carry  full  rated  current  at  rated  voltage  and 
frequency.  Their  output  under  this  condition  would  be  zero  if 
the  current  and  voltage  were  out  of  phase  by  90  degrees.  It  will 
be  a  maximum  when  the  current  and  voltage  are  in  phase.  This 
latter  condition  is  most  to  be  desired  and  is  approached  in  prac- 
tice, although  seldom  exactly  attained.  The  common  unit 
for  apparent  power  for  large  power  apparatus  is  the  kilovolt- 
ampere  (abbreviated  kv-a.).  This  bears  the  same  relation  to  the 
volt-ampere  as  the  kilowatt  bears  to  the  watt. 

Power-factor. — For  steady  voltages  and  currents  the  power  is 
always  given  by  the  product  of  volts  and  amperes,  that  is  by  the 
volt-amperes.  For  alternating  currents  this  is  true  only  when 
the  voltage  and  current  are  in  phase.  In  general,  for  sinusoidal 
waves,  power  is  (equation  (8))  El  cos  6.  It  is  equal  to  the  volt- 
amperes  multiplied  by  a  factor,  cos  0,  which  is  determined  by  the 
phase  relation  of  the  voltage  and  current.  This  factor,  cos  0, 
is  called  the  power-factor  of  the  circuit.  When  the  current  and 
voltage  waves  are  not  sinusoidal,  the  factor  cos  6  has  no  significance 
except  when  considered  with  respect  to  the  equivalent  sine  waves. 
These  will  be  explained  later. 

Theoretically,  the  power-factor  may  have  any  value  from  zero 
to  unity,  both  inclusive.  Although  it  is  possible  to  obtain  zero 
power-factor  experimentally  by  the  use  of  a  synchronous  motor 
which  receives  enough  power  mechanically  to  supply  its  losses, 
zero  power-factor  is  not  reached  in  practice.  It  would  be  a 
very  undesirable  condition  in  most  cases.  Unity  power-factor 
may  be  obtained  and  is  not  uncommon  in  alternating-current 
work. 

The  power-factor  of  a  circuit  is  always  the  ratio  of  the  true 


POWER  57 

power  to  the  apparent  power,  i.e.,  the  ratio  of  the  watts  to  the 
volt-amperes.  It  is  merely  a  ratio  and  therefore  independent  of 
the  units  used  except  that  the  true  power  and  the  apparent  power 
must  be  expressed  in  the  same  system  of  units.  Power-factor 
is  commonly  expressed  in  per  cent,  although  it  is  frequently  given 
as  a  fraction.  It  must  always  be  taken  as  a  fraction  when  used 
in  numerical  expressions.  Power-factor  is  the  cosine  of  an  angle 
only  for  sinusoidal  waves  of  current  and  voltage. 

Power-factor  =  ,T  ,,  (10) 

Volt-amperes 

This  expression  for  power-factor  holds  regardless  of  wave  form. 
Reactive-factor. — The  expression 

sin  6  =  V  1  --  (cos  0)2 

=  sin  [cos"1  (Power-factor)]    (11) 

is  called  the  reactive-factor.  It  is  true,  however,  only  for 
sinusoidal  waves.  When  the  waves  are  not  sinusoidal,  the  re- 
active-factor is  given  by 

Reactive-factor  =  \/l  —  (Power-factor)2  (12) 

The  reactive-factor  is  of  considerable  importance,  since  a 
knowledge  of  its  magnitude  is  necessary  when  determining  the 
size  of  the  apparatus  required  to  raise  the  power-factor  of  a  circuit 
to  unity,  or  by  any  desired  amount.  The  reactive-factor  is 
positive  for  lagging  currents  and  negative  for  leading  currents 
when  6  is  the  lag  angle  of  the  current  with  respect  to  the  voltage. 

Measurement  of  Average  Power. — The  average  power  in  a 
circuit  may  be  measured  by  an  electrodynamometer.  If  the  fixed 
coil  of  such  an  instrument  is  placed  in  series  with  one  main  of  the 
circuit  in  which  the  power  is  to  be  measured  and  its  movable 
coil  is  connected  in  series  with  a  large  non-inductive  resistance 
and  then  shunted  between  the  two  mains,  the  average  torque 
developed  between  the  two  coils  will  be  proportional  to  the  aver- 
age product  of  the  instantaneous  values  of  current  and  voltage. 
The  torque  exerted  between  two  coils  which  are  free  to  turn  about 
a  common  diameter  is  proportional  to  the  product  of  the  currents 
they  carry.  In  the  electrodynamometer  used  as  a  wattmeter, 
the  fixed  coil  (usually  called  the  current  coil)  carries  line  current. 
The  moving  coil  (usually  called  the  potential  coil),  in  series 


58  PRINCIPLES  OF  ALTERNATING  CURRENTS 

with  non-inductive  resistance,  carries  a  current  which  is  pro- 
portional to  the  voltage  across  its  terminals,  provided  the  non- 
inductive  resistance  is  large  compared  with  the  inductance  of  the 
potential  windings  of  the  instrument. 


Average  torque  =  — 


i  f  ?  • 

-*J," 

-  r  Pi 

J  Jo   /t 


-.' 

where  ^  and  *c  are,  respectively,  the  instantaneous  currents 
carried  by  the  potential  and  current  coils,  ft  is  the  non-inductive 
resistance  in  series  with  the  potential  coil  plus  the  resistance  of 
the  potential  coil  and  ep  is  the  instantaneous  voltage  across  the 
terminals  of  the  potential  coil  circuit.  P  is  the  average  power. 
The  moving  coil  will  deflect  until  the  average  torque  developed 
in  it  is  just  balanced  by  the  torsion  of  the  control  springs  attached 
to  its  shaft.  By  proper  calibration  the  instrument  may  be  made 
to  indicate  the  average  power.  When  the  power  in  high-voltage 
circuits  is  to  be  measured,  it  is  necessary  to  use  transformers  with 
both  the  current  and  potential  coils  of  the  instrument. 

Measurement  of  Power-factor. — Although  there  are  special 
instruments  which  will  indicate  the  power-factor  of  the  circuit  in 
which  they  are  connected,  they  are  not  very  accurate  and  are 
not  in  very  general  use  except  on  switchboards.  The  power- 
factor  of  a  circuit  is  generally  determined  by  measuring  the 
voltage,  current  and  power  by  means  of  a  voltmeter,  an  ammeter 
and  a  wattmeter  and  then  taking  the  ratio  of  the  watts  to  the 
volt-amperes.  The  reactive-factor  is  calculated  from  the  power- 
factor  by  equation  (12),  page  57. 

Energy  and  Wattless  or  Quadrature  Components  of  Current. 
When  dealing  with  sinusoidal  waves,  it  is  often  convenient  to 
resolve  the  current  into  two  right-angle  components,  one  in  phase 
Awith  the  voltage,  the  other  in  quadrature  with  it.  These  two 
'components  are  I  cos  0  and  I  sin  0.  The  reason  for  selecting 
; these  two  components  will  be  made  clear  by  what  follows.  Let 

e  =  Em  sin  cat 

.i  =  Im  sin  (cat  -  Q) 


POWER  59 

Then  since  i  =  Im  cos  0  sin  cot  —  Im  sin  0  cos  at 
p  =  Em  sin  cot  (Im  cos  0)  sin  cot  —  Em  sin  cot  (7TO  sin  0)  cos  cot  (13) 

i  r  * 

P  =  ^  I   Em  sin  cot  (!„  cos  0)  sin  cotdt 
7  Jo 

i  rr 

—  —  I   Em  sin  cot  (7ro  sin  0)  cos  cotdt  (14) 

i  Jo 

The  second  integral  vanishes  and  therefore  contributes  nothing 
to  the  average  power  or  watts  of  the  circuit.  The  power  may  be 
regarded  as  due  to  a  voltage  e  =  Em  sin  cot  and  a  current  i  = 
(Im  cos  0)  sin  cot,  whose  maximum  values  are  Em  and  Im  cos  6 
respectively  and  whose  phase  difference  is  zero.  Since  the  power 
is  contributed  wholly  by  the  component  of  the  current  Im  cos  0, 
which  is  in  phase  with  Em,  this  component  is  known  as  the 
energy,  power  or  active  component  of  the  current,  or  simply  the 
energy,  power  or  active  current.  The  same  terms  are  applied  to 
the  corresponding  component  of  the  effective  current.  In 
fact  when  energy,  power  or  active  current  is  mentioned,  the 
energy,  power  or  active  component  of  the  effective  current  is 
understood. 

The  second  integral  may  be  regarded  as  representing  the  power 
due  to  a  voltage  e  =  Em  sin  cot  and  a  current  i  =  —  (Im  sin  0}  cos  cot 
or  i  —  (Im  sin  0)  sin  (cot  —  90°),  whose  maximum  values  are 
respectively  Em  and  Im  sin  0  and  whose  phase  difference  is  90 
degrees.  It  has  already  been  shown  that  when  a  voltage  and 
a  current  are  in  quadrature  the  average  power  is  zero.  The 
component  Im  sin  0  of  the  current  contributes  nothing  to  the 
average  power  of  the  circuit  and  is  therefore  known  as  the  quad- 
rature, wattless  or  reactive  component  of  the  current  or  simply  the 
quadrature,  wattless  or  reactive  current.  The  same  terms  are 
applied  to  the  corresponding  component  of  the  effective  current. 
Although  the  component  Im  sin  0  contributes  nothing  to  the 
average  power  of  the  circuit,  its  presence  increases  the  resultant 
current  and  therefore  increases  the  copper  loss  for  a  given  amount 
of  power.  The  resultant  current  is  always  equal  to  the  square 
root  of  the  sum  of  the  squares  of  the  active  and  the  reactive  com- 
ponents of  the  current.  The  resultant  current  causes  the  copper 
loss  but  only  its  active  component  contributes  to  the  average 
power. 


60  PRINCIPLES  OF  ALTERNATING  CURRENTS 

If  E  and  /  represent  the  root-mean-square  values  of  the  voltage 
and  current,  the  expression 

Pa  =  El  cos  6  =  El  X  power  factor 

represents  the  average  power  due  to  the  active  or  energy  com- 
ponent of  the  current.  Pa  is  the  true  average  power  of  the 
circuit.  It  is  sometimes  called  the  active  power  or  the  active 
volt-amperes,  i.e.,  the  volt-amperes  due  to  the  active  component 
of  the  current. 

The  second  term  of  the  second  member  in  equation-  (14)  is  the 
average  power  due  to  the  wattless  or  reactive  component  of  the 
current.  This  average  is  zero.  The  expression  for  the  instan- 
taneous power  due  to  the  reactive  component  of  the  current  is 

pr  =  —Emlm  sin  0  sin  ut  cos  ut 

=  -^^  sin  0  sin  2co£  (15) 


This  has  double  frequency  and  represents  an  oscillation  of 
power  between  the  generating  source  and  the  load,  of  maximum 
value 

Emlm  sin  0  =  El  sin  0  (16) 


2 

Its  average  value  over  one  complete  loop,  i.e.,  over  one  quarter 
of  a  period  of  the  current  wave,  is 

-  El  sin  0 

7T 

This  expression  is  of  no  particular  use  in  practice. 
In  addition  to  representing  the  maximum  value  of  the  power 
oscillation  caused  by  the  reactive  component  of  the  current. 

Tji      j 

expression  (16),  i.e.,  -^~  sin  0  =  El  sin  0,  also  represents  the 

root-mean-square  volt-amperes  due  to  the  reactive  component 
of  the  current.  It  is  sometimes  called  the  reactive  power.  A 
better  name  is  reactive  volt-amperes,  i.e.,  the  volt-amperes  due 
to  the  reactive  component  of  the  current.  Reactive  power  or 
reactive  volt-amperes  is  the  product  of  the  total  volt-amperes 
and  the  reactive-factor. 

The  total  volt-amperes  of  a  circuit  are  equal  to  the  square 


POWER 


61 


root  of  the  sum  of  the  squares  of  the  active  and  reactive  powers. 


Volt-amperes  =  V(E  I  cos  0)2  +  (E  I  sin  0)2 


=  V(Active  power)2  -f  (Reactive  power)2 
=  EI 


(17) 


It  follows  from  equation  (17)  that  with  sinusoidal  waves, 
power-factor  may  be  defined  as  the  ratio  of  the  active  power  to 
the  square  root  of  the  sum  of  the  squares  of  the  active  and 
reactive  powers. 

Although  the  average  value  of  the  oscillation  of  power  caused 
by  the  reactive  component  of  the  current  is  zero,  the  oscillation 
cannot  take  place  without  producing  a  copper  loss.  The  oscilla- 
tion cannot  contribute  to  the  available  power,  but  it  does  increase 
the  copper  loss  in  a  circuit.  For  this  reason  it  is  desirable  to 
operate  circuits  at  as  nearly  unity  power-factor  as  possible. 

The  active  or  energy  and  reactive  or  wattless  components  of 
current  for  a  circuit  carrying  /  amperes  at  a  power-factor  cos  0  are 

Active  current  =  I  cos  0  =  /  X  Power-factor  (18) 

Reactive  current  -=  I  sin  6  =  1  Vl  —  cos2  9 

=  I  Vl  -  (Power-factor)2 

=  /  X  Reactive-factor  (19) 

For  sinusoidal  waves,  the  watts  will  always  be  given  by  the 
product  of  volts  and  energy  amperes. 


t-o 


FIG.  20. 


Curves  of  voltage,  active  and  reactive  components  of  the 
current  and  the  power  curves  corresponding  to  the  two  compo- 
nents of  the  current  are  shown  in  Fig.  20.  The  two  power  curves 


62  PRINCIPLES  OF  ALTERNATING  CURRENTS 

are  obtained  by  plotting  the  two  terms  of  the  second  member 
of  equation  (14)  separately  against  time.  The  two  component 
currents  Im  cos  0  and  Im  sin  6  are  assumed  to  be  equal.  This  cor- 
responds to  a  power-factor  of  0.707  and  a  phase  angle  of  45  degrees 
between  voltage  and  current.  The  phase  angle  is  assumed  to 
be  an  angle  of  lag  of  the  current  with  respect  to  the  voltage. 

The  power  curves  corresponding  to  both  components  of  the 
current  are  double  frequency  curves  and  represent  a  periodic 
flow  of  power.  The  power  due  to  the  energy  component  of  the 
current,  Im  cos  6,  is  always  positive  and  represents  the  power 
given  out  by  the  circuit.  The  power  due  to  the  reactive  compo- 
nent of  the  current,  Im  sin  6,  alternates  between  plus  and  minus, 
and  the  average  power  is  zero.  The  combined  effect  of  the  two 
components  produces  a  power  curve  partly  above  and  partly 
below  the  axis  of  time  similar  to  that  shown  in  Fig.  19,  page  54. 

At  certain  instants  the  resultant  power  developed  is  positive 
and  at  others  it  is  negative  (see  Fig.  19).  The  average  power 
is  found  by  subtracting  the  area  enclosed  between  the  axis  of 
time  and  the  negative  loops  from  the  area  enclosed  between  this 

axis  and  the  positive  loops  and  dividing  the  remainder  by  T  =  j- 

The  resultant  area  is  the  same  as  the  area  enclosed  between  this 
axis  and  the  power  curve  corresponding  to  the  energy  component 
of  the  current,  Im  cos  6. 

The  conditions  shown  in  Figs.  17  and  18  represent  limiting 
cases.  In  the  first,  the  reactive  component  of  the  current  is  zero. 
In  the  second,  the  active  component  is  zero. 

It  is  important  to  remark  that  the  power  in  a  single-phase 
circuit  is  always  fluctuating,  becoming  zero  or  even  negative  at 
certain  instants  of  time.  This  means  that  a  circuit  may  supply 
power  to  the  generator  during  part  of  a  cycle.  The  conditions 
are  different  in  a  balanced  polyphase  circuit.  Here  the  deficiency 
of  power  at  any  instant  in  one  phase  is  made  up  by  an  excess  of 
power  in  the  other  phases.  The  total  power  of  such  a  circuit,  i.e., 
the  resultant  power  for  all  the  phases,  does  not  fluctuate. 

Energy  and  Wattless  or  Quadrature  Components  of  the  Volt- 
age.— Instead  of  dividing  the  current  into  energy  and  wattless 
components  with  respect  to  the  voltage,  the  voltage  may  be 
similarly  divided  into  energy  and  wattless  components  with  re- 
spect to  the  current. 


POWER  63 

The  two  components  of  the  voltage  are 

ea  =  (Em  cos  0)  sin  ut  (20) 

er  =  -(Em  sin  0)  cos  at  =  (Em  sin  0)  sin  (at  -  90°)     (21) 

(See  equation  (13))  In  this  case  the  power  may  be  considered  to 
be  due  to  a  voltage  e  =  (Em  cos  0)  sin  ut  and  a  current  i  =  Im  sin 
orf  in  phase  with  it.  The  component  e  =  —  (Em  sin  0)  cos  ut  of 
the  voltage  is  in  quadrature  with  the  current  i  =  Im  sin  ut  and 
therefore  produces  no  average  power.  The  component  Em  cos  0 
is  known  as  the  energy,  power  or  active  component  of  the  volt- 
age. The  component  Em  sin  0  is  known  as  the  wattless, 
quadrature  or  reactive  component  of  the  voltage. 

Measurement  of  Reactive  Power,  i.e.,  Reactive  Volt-amperes. 
The  reactive  power  of  a  circuit  may  be  measured  by  an  elec- 
trodynamometer  connected  in  the  same  way  as  for  measuring 
true  power,  provided  the  current  through  its  potential  circuit 
can  be  made  to  lag  the  voltage  across  its  terminals  by  90  degrees. 
For  measurement  of  true  power,  the  current  in  the  potential 
circuit -must  be  in  phase  with  the  voltage  across  its  terminals. 
The  necessary  lag  of  90  degrees  for  the  current  in  the  potential 
coil  may  be  obtained  by  connecting  it  in  series  with  a  large 
inductance  instead  of  a  large  non-inductive  resistance  as  is 
done  for  power  measurements.  (See  page  57.)  Due  to  the 
fact  that  an  inductance  cannot  be  made  without  some  resistance, 
the  current  in  the  potential  coil  will  not  lag  by  exactly  90  degrees 
if  the  series  inductance  alone  is  used.  It  may,  however,  be 
made  to  lag  exactly  90  degrees  by  shunting  the  potential  coil  with 
a  suitable  non-inductive  resistance.  Since  the  effect  of  induc- 
tance depends  upon  frequency,  a  reactive-power  meter  will  indi- 
cate El  sin  0  only  when  the  frequency  is  that  for  which  the 
instrument  is  adjusted.  The  effect  of  the  inductance  and 
shunted  resistance  will  be  understood  after  studying  Chapter  VII. 

The  average  torque  developed  between  the  coils  of  the  electro- 
dynamometer  will  be 

Average  torque  =  ~  I    M«<B 

1   CT 
=  =  I    kEm  sin  (at  -  90°)/TO  sin  (  ut-  0)dt 


64  PRINCIPLES  OF  ALTERNATING  CURRENTS 

where  Em  and  Im  are  the  maximum  values  of  the  voltage  and 
current  of  the  circuit  and  k  is  a  constant  of  proportionality 
between  the  voltage  Em  sin  cot  across  the  potential  circuit  and  its 
current  ip.  This  assumes  a  pure  inductive  circuit. 

Tf>       T 

Average  torque  =  k  — 7™  sin  6 
=  kEI  sin  0 
=  k  X  (Reactive  power) 

By  suitable  calibration  the  instrument  may  be  made  to  indicate 
reactive  power  directly. 

Vectors  Representing  Voltage  Rise  and  Voltage  Fall. — When 
plotting  vector  diagrams  and  in  making  computations  involving 
currents  and  voltages,  it  is  always  necessary  to  distinguish 
between  voltage  rise  and  voltage  fall  or  drop. 

If  the  direction  ab  in  any  circuit  is  taken  positive  for  current 
flow,  the  current  is  actually  positive  and  actually  flows  from  a 
to  b  when  the  revolving  vector  representing  the  current  7^  lies 
in  the  first  or  second  quadrants.  It  is  actually  negative  and  flows 
from  b  to  a  when  the  revolving  vector  representing  it  lies  in  the 
third  and  fourth  quadrants.  If  ab  is  taken  as  the  positive  direc- 
tion for  the  current  vector,  Vab,  representing  a  voltage  associated . 
with  the  current  7a&,  is  a  voltage  rise,  if  on  the  average  it  is 
actually  increasing  in  the  direction  ab,  i.e.,  there  is  an  actual  rise 
in  the  direction  ab,  when  the  current  vector  lies  in  the  first  and 
second  quadrants.  In  other  words,  if  the  direction  ab  is  assumed 
positive  for  the  current  flow  7«&,  the  voltage  Vab  is  a  voltage  rise 
and  is  considered  positive  when  its  active  component,  i.e.,  its 
component  in  phase  with  the  current,  is  an  actual  rise  in  voltage 
in  the  direction  ab  when  the  current  flow  is  positive.  Vab  is  a 
voltage  drop  or  fall  and  is  negative  when  its  active  component, 
i.e.,  its  component  in  phase  with  the  current,  is  an  actual  drop 
in  voltage  in  the  direction  ab  when  the  current  flow  is  from 
a  to  b. 

A  voltage  vector  may  be  expressed  either  as  a  voltage  rise  or  a 
voltage  fall.  If  it  represents  a  rise  in  voltage  and  power  is 
absorbed,  its  component  in  phase  with  the  current,  i.e.,  its  active 
component,  will  be  negative  with  respect  to  the  current  vector.  In 
this  case  the  voltage,  on  the  average,  is  actually  a  voltage  fall  in 


POWER  65 

the  direction  of  the  current  flow.  On  the  average  there  is  actually 
a  voltage  drop  in  the  direction  of  current  flow.  However,  the 
voltage  may  be  considered  to  be  actually  rising  in  a  negative 
direction  with  respect  to  the  current. 

If  a  vector  Vab  is  considered  .to  represent  a  voltage  rise  in 
the  direction  ab,  there  is  an  actual  rise  in  voltage  between  a  and  b 
when  Vab  lies  in  the  first  and  second  quadrants.  There  is  an 
actual  fall  in  voltage  in  the  direction  ab  when  Vab  lies  in  the  third 
and  fourth  quadrants.  If  V^  represents  a  voltage  rise  in  the 
direction  ab,  —Vab  is  a  negative  rise  in  the  direction  ab  or  is  a 
drop  in  the  direction  ab.  There  is  an  actual  drop  in  voltage  in  the 
direction  ab  when  the  vector  —  Fa&  lies  in  the  first  and  second 
quadrants  and  an  actual  rise  in  the  direction  ab  when  it  lies  in 
the  third  and  fourth  quadrants. 

In  general,  when  only  power  delivered  or  both  power  delivered 
and  received  are  to  be  considered  in  the  same  problem,  it  is  best 
to  let  a  positive  vector  represent  a  voltage  rise.  A  negative  vec- 
tor will  then  represent  a  voltage  fall  or  drop.  According  to  this 
convention  a  vector  V  would  represent  a  voltage  rise  in  a  circuit 
in  the  direction  which  is  assumed  positive  for  the  current.  A 
vector  —  V  would  be  the  voltage  drop  in  the  same  direction  or  a 
rise  in  the  opposite  direction.  However,  as  has  been  stated, 
when  dealing  only  with  power  absorbed  it  is  convenient  and 
customary  to  consider  that  a  positive  vector  represents  a  voltage 
drop.  When  this  latter  convention  is  adopted,  V  would  represent 
a  voltage  drop  in  the  direction  in  which  the  current  is  considered 
positive  and  —  V  would  represent  a  voltage  rise  in  the  same 
direction. 

When  a  voltage  rise  is  considered  positive,  power  delivered 
is  positive  and  power  absorbed  is  negative.  When  a  voltage 
drop  is  considered  positive,  power  absorbed  is  positive  and  power 
delivered  is  negative. 

The  mere  reversal  of  an  alternating  current  or  voltage  during 
a  cycle  does  not  change  the  sign  of  the  vector  representing  it. 
If  a  voltage  vector  represents  a  voltage  rise,  i.e.,  its  component  in 
phase  with  the  current  is  rising  in  a  positive  direction  when 
the  current  flow  is  positive,  it  will  still  be  a  voltage  rise  when  its 
position  has  reversed  due  to  its  rotation,  since  the  position  of  the 
current  vector  has  also  reversed.  Its  component  in  phase  with 
the  current  will  still  represent  a  voltage  rise  in  the  direction  of  cur- 


66 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


rent  flow.  It  will  be  rising  in  a  negative  direction  with  respect 
to  the  direction  assumed  positive  for  current  flow,  but  the  cur- 
rent flow  will  also  have  reversed  and  will  be  negative. 

Expression  for  Power  when  the  Voltage  and  Current  are  in 
Complex. — Let 

E  =  e  +  je' 


FIG.  21. 


where  E  and  I  are  the  effective  or 
root-mean-square     values    of    the 
voltage   and   current.     The   small 
letters    without    primes     are     the 
real  components;  with  primes  they 
represent     the     imaginary      com- 
ponents.    The  two  vectors,  E  and 
'x  /,   are  shown  in  Fig.  21. 
From  Fig.  21 


P  =  El  cos  6 

=  El  cos  (0i  -  02) 
=  EI(cos  0i  cos  02 


sin  B\  sin 


ei  •+  e'i' 


(22) 


The  component  e  of  the  voltage  cannot  produce  power  with  the 
component  i'  of  the  current,  since  e  and  z'-are  in  phase  quadra- 
ture. Neither  can  the  component  e'  of  the  voltage  produce 
power  with  the  component  i  of  the  current,  since  they  are  also  in 
phase  quadrature.  The  components  in  the  products  ei  and  e'i' 
are  in  phase.  They  are  the  only  components  that  can  contribute 
to  the  power. 

The  expression  e'i  —  ei'  is  the  reactive  power  or  reactive 
volt-amperes.  This  may  be  shown  as  follows: 

Reactive  power  =  Pr  =  E I  sin  0 

=  EI  sin  (0!  -  02) 

=  EI(sm  0i  cos  02  —  cos  0i  sin  02) 


POWER  67 

When  0  is  taken  as  the  angle  of  lag  of  the  current  with  respect 
to  the  voltage,  the  current  will  lag  the  voltage  when  the  reactive 
power  is  positive. 

E  I  =  V  (active  volt-amperes)2  +  (reactive  volt-amperes)2 

=  V  (true  power)2  +  (reactive  power)2 
=  volt-amperes. 

It  should  be  noted  that  the  power  P  =  ei  +  di'  is  not  given 
by  the  product  of  the  complex  values  of  current  and  voltage. 
This  product  is  evidently  (ei  —  e'i')  +  j(ei'  +  e'i)  and  has  no 
significance. 

Example  of  the  Calculation  of  Power,  Power -factor,  etc. — Let 
an  alternating  voltage  E  =  100  +  J50  be  impressed  on  a  circuit 
whose  constants  are  such  that  the  current  resulting  is  /  =  20  — 
;30.  E  and  /  are  both  root-mean-square  or  effective  values. 

E  =  V(100)2  -I-  (50)2  =  111.8  volts 
7  =  V(20)2  +  (30)2    =  36.06  amperes 

Let  BE  be  the  phase  angle  of  E  with  respect  to  the  reference 
axis  and  let  0/  be  the  corresponding  angle  for  the  current. 

tan  BE  =  ~   =  0.5  BE  =  +26.57  degrees. 

QQ 

tan  B,  =  -=JT-  =  -1.5  fl/  =  -56.31  degrees. 

20 

The  angle  of  lag  of  the  current  with  respect  to  the  voltage 
is  then 

Oi  =  (+26.57)  -  (-56.31)  =  82.88  degrees. 
The  current  /  lags  the  voltage  by  82.88  degrees. 
Volt-amperes  =  111.8  X  36.06  =  4030 

Power  =  100  X  20  +  50  X  (-30) 
=  500  watts. 

p 

Power-factor  =  -^j  =  cos  B*E 

500        0.1240 


4030 

BrE  =  cos-1  (0.1240)  =  82.88  degrees. 
Reactive-factor  =  sin  0*  =  0.9923 


68  PRINCIPLES  OF  ALTERNATING  CURRENTS 

In  finding  the  reactive-factor,  0J,  is  taken  as  positive  since  it 
represents  a  lag  of  the  current  behind  the  voltage. 

Reactive  power  =  E  I  sin  OrE 

=  4030  X  0.9923 

=  3999     reactive    watts     or    reactive    volt- 
amperes. 

Active  component  of  the  current       =  Ia  =  I  cos  Q'E 

=  36.05  X  0.1240 
=  4.47  amperes. 

Reactive  component  of  the  current  =  Ir  =  /  sin  B1E 

=  36.05  X  0.9923 
=  35.78  amperes. 


CHAPTER  IV 

NON-SINUSOIDAL  WAVES 

Wave  Form  of  Alternators. — The  wave  form  of  commercial 
alternators  is  never  exactly  sinusoidal  and  under  certain  condi- 
tions it  may  differ  considerably  therefrom.  The  armature 
windings  of  multipolar  alternators  consist  of  as  many  identical 
elements  per  phase  as  there  are  pairs  of  poles.  These  elements 
are  spaced  180  electrical  degrees  apart  on  the  armature  and  are 
usually  connected  in  series.  Polyphase  alternators  have  as 
many  groups  of  windings  as  there  are  phases.  These  are  dis- 

360 
placed  from  one  another  by  -  -  electrical  degrees,  where  n  is  the 

71 

number  of  phases.  Since  the  groups  of  windings  are  always 
identical,  the  voltages  induced  in  them  will  be  equal  in  magnitude 

360  1 

but  will  be  displaced  in  time  phase  by  -   -  degrees  or  by  -  of 

the  time  of  a  complete  cycle.  The  windings  of  a  polyphase 
alternator  are  always  interconnected  in  star  or  in  mesh  (See 

hapt.  VIII)  in  order  to  diminish  the  number  of  terminals.  In 
general,  except  for  single  phase,  there  are  as  many  terminals  as 
there  are  phases.  Most  commercial  alternators  are  three-phase, 
although  two-phase  or  four-phase  alternators  are  occasionally 
used. 

If  the  sides  of  the  coils  of  an  armature  winding  are  180  electrical 
degrees  apart  and  if  one  side  of  a  coil  occupies  any  given  position 
with  respect  to  a  north  pole,  the  other  side  of  the  coil  will  occupy 
an  exactly  similar  position  with  respect  to  the  adjacent  south 
pole.  If  the  poles  are  similar  so  that  the  flux  distribution  pro- 
duced by  each  is  the  same,  the  voltages  induced  in  the  two  coil 
sides  will  be  equal  in  magnitude  but  opposite  in  direction.  That 
is,  if  one  acts  from  the  front  to  the  back  of  the  coil,  the  other  acts 
Tom  the  back  to  the  front.  At  every  instant,  the  voltage  gene- 
rated in  the  coil  will  be  the  difference  of  the  voltages  generated 
n  its  two  sides. 


70  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  voltage  generated  in  a  coil  side  is  equal  to 

e  =  &LsZ  X  10-8  (1) 

where  L  is  the  effective  length  of  the  coil  side,  i.e.,  the  length 
of  the  part  cutting  the  flux,  and  Z  is  the  number  of  inductors  in 
each  coil.  An  inductor  is  one  of  the  two  active  sides  of  each 
turn  of  an  armature  coil.  Each  turn  has  two  inductors.  The 
inductors  are  the  parts  of  an  armature  coil  which  cut  flux.  (B 
is  the  flux  density  in  gausses  at  the  inductor  and  s  is  the  velocity 
of  the  inductor  in  centimeters  per  second. 

The  velocity  s  is  fixed  by  the  radius  of  the  armature,  the  fre- 
quency and  the  number  of  poles.  The  only  variable  is  the  flux 
density  (B  over  the  pole  face.  This  is  determined  by  the  total  flux 
per  pole  and  the  shape  of  the  pole  face.  Since  the  only  variable  is 
the  flux  density,  the  shape  of  the  voltage  wave  must  be  the  same 
as  the  shape  of  the  curve  of  flux  distribution  in  the  air  gap. 
If  the  pole  faces  are  concentric  with  the  armature  surface,  the 
flux  distribution  in  the  air  gap  will  be  nearly  constant  over  the 
pole  faces  and  will  drop  rapidly  to  zero  at  points  midway  between 
poles.  If  the  edges  of  the  poles  are  chamfered  so  that  the  air 
gap  at  the  edges  of  the  pole  is  longer  than  at  the  middle,  the  flux 
density  will  no  longer  be  uniform  over  the  pole  face  but  will  be 
greatest  at  the  centre.  By  properly  shaping  the  pole  face,  the! 
flux  density  may  be  made  to  vary  approximately  sinusoidally  in 
passing  from  a  point  midway  between  any  pair  of  poles  to  a 
point  midway  between  the  next  pair.  If  it  could  be  made 
exactly  sinusoidal,  the  voltage  generated  in  each  coil  side  would 
also  be  sinusoidal,  since  it  is  proportional  to  the  flux  density  at 
each  instant.  If  the  pole  faces  are  nearly  concentric  with  the 
armature,  the  total  flux  for  a  given  permissible  maximum  flux 
density  will  be  greater  than  when  the  pole  faces  are  chamfered, 
but  the  voltage  wave  will  be  flat.  If  the  pole  faces  are  chamfered 
too  much,  the  wave  form  will  be  peaked,  i.e.,  more  peaked  than 
a  sinusoid. 

Even  if  the  poles  of  an  alternator  could  be  shaped  to  give 
an  exactly  sinusoidal  flux  distribution  at  no  load,  the  wave  form 
of  the  alternator  under  load  might  differ  considerably  from  a 
sinusoid,  due  to  the  effect  of  armature  reaction  in  distorting  the 
flux  distribution.  In  general,  except  when  the  current  is  in.; 


NON-SINUSOIDAL  WAVES  71 

quadrature  with  the  induced  voltage,  armature  reaction  tends  to 
crowd  the  flux  towards  one  side  of  each  pole  and  to  give  rise  to  a 
distorted  voltage  wave  which  may  differ  greatly  from  a  sinusoid. 
Although  the  flux  distribution  may  become  badly  distorted 
under  load,  it  is  possible  to  so  arrange  an  armature  winding  that 
the  voltage  between  its  terminals  will  still  be  nearly  sinusoidal. 

Nearly  all  alternators  have  distributed  armature  windings. 
That  is,  the  coils  for  any  one  phase  are  distributed  among  several 
pairs  of  slots  per  pair  of  poles  instead  of  being  placed  in  a  single 
pair  of  slots  per  pair  of  poles.  In  this  case  the  voltage  induced  in 
the  coils  will  not  be  in  phase  but  will  differ  in  phase  by  an  angle 
equal  to  the  angle  in  electrical  degrees  between  the  slots.  Also, 
the  coil  pitch  is  often  less  than  180  electrical  degrees,  i.e.,  the 
coil  sides  are  less  than  180  electrical  degrees  apart.  A  very 
common  pitch  is  %  X  180  =  150  electrical  degrees. 

One  important  effect  of  distributing  a  winding  and  also 
shortening  its  pitch  is  to  smooth  out  the  wave  form  and  make  it 
more  nearly  sinusoidal  by  reducing  or  eliminating  the  components 
in  the  voltage  which  cause  it  to  differ  from  a  sinusoid.  As  will 
be  explained  later,  any  distorted  voltage  wave  may  be  resolved 
into  a  series  of  sinusoidal  components  having  frequencies  which  are 
1,  3,  5,  7,  9,  etc.  times  the  fundamental  frequency,  i.e.,  the 
frequency  determined  by  the  speed  and  the  number  of  poles. 

If  the  angle  between  the  slots  is  a  electrical  degrees,  the 
voltage  generated  in  coils  which  are  in  adjacent  slots  will  be 
out  of  phase  by  a  degrees  for  the  component  of  fundamental 
frequency,  3a  degrees  for  the  component  of  third  frequency, 
5a  for  the  component  of  fifth  frequency,  etc.  The  effect  of 
this  difference  in  phase  is  to  reduce  the  distorting  components 
in  the  voltage  wave  and  thus  to  make  the  voltage  more  nearly 
sinusoidal.  A  similar  effect  is  produced  by  shortening  the 
pitch. 

By  the  use  of  a  suitably  shortened  pitch,  any  one  component  in 

the  voltage  wave  may  be   completely  eliminated  in  the   coil 

4 

voltage.     For  example,  if  the  coil  sides  are  >  X  180  =  144  elec- 

o 

trical  degrees  apart,  the  components  of  fifth  frequency  in  the  two 
sides  of  any  coil  will  be  144  X  5  =  720  degrees  out  of  phase. 
They  will  be  displaced  two  whole  wave  lengths  and  will  there- 


72  PRINCIPLES  OF  ALTERNATING  CURRENTS 

fore  be  in  phase.     Since  the  coil  voltage  is  the  vector  difference 

4 
between  the  voltages  generated  in  the  two  coil  sides,  a  ^  pitch 

will  completely  eliminate  the  voltage  component  of  fifth  fre- 
quency. Any  component  may  also  be  eliminated  by  distributing 
the  winding  among  a  suitable  number  of  properly  spaced  slots. 

The  triple  frequency  component  in  the  voltage  is  eliminated 
in  a  three-phase  alternator  by  the  inter-connection  of  the  phases. 

Instead  of  trying  to  completely  eliminate  certain  of  the  dis- 
torting components  by  the  use  of  a  suitable  coil  pitch  coupled 
with  distribution  of  the  winding,  a  pitch  and  distribution  are 
usually  chosen  which  considerably  reduce  a  number  of  the 
distorting  components  without  completely  eliminating  any  one 
of  them.  Better  average  wave  form  may  be  obtained  in  this 
way  than  when  certain  components  are  completely  eliminated. 

Although  it  is  possible  to  design  an  alternator,  by  the  use 
of  shortened  pitch,  a  distributed  winding  and  certain  other 
devices,  to  give  practically  a  sinusoidal  wave  of  voltage  under 
load  as  well  as  at  no  load,  alternators  are  not  commercially  so 
designed  except  when  they  are  to  be  used  for  special  work,  such  as 
cable  testing,  which  requries  sinusoidal  voltage.  Under  ordinary 
conditions,  the  wave  form  of  commercial  alternators  does  not  differ 
greatly  from  a  sinusoid.  Even  though  the  voltage  wave  of  an 
alternator  were  exactly  sinusoidal  under  all  conditions,  the 
current  wave  would  not  necessarily  be  sinusoidal.  It  is  there- 
fore necessary  to  consider  the  conditions  existing  when  the  voltage 
and  current  waves  are  not  simple  harmonic  functions  of  the  time. 

Representation  of  a  Non- sinusoidal  Current  or  Voltage 
Wave  by  a  Fourier  Series. — Any  single-valued  periodic  function 
may  be  resolved  into  a  Fourier  series  consisting  of  sine  and  cosine 
terms  of  different  relative  magnitudes  and  with  frequencies  which 
are  in  the  ratios  of  1,  2,  3,  4,  5,  etc.  to  the  frequency  of  the 
fundamental  period  of  the  function.*  For  example,  any  single- 
valued  periodic  quantity,  like  the  voltage  or  current  of  an  alter- 
nator, may  be  represented  by  the  following  series.  For  the 
alternator  AQ  is  zero. 

e  =  AQ  +  AI   sin  ut  +  BI   cos  ut  +  A2   sin   2cof  +  B2   cos   2ut 
+  A3  sin  3orf  +  53  cos  3otf  +  etc.  (2) 

*  See  Fourier's  Series  and  Spherical  Harmonics,  Byerly. 


NON-SINUSOIDAL  WAVES  73 

The  sine  and  cosine  terms  are  called  harmonics.  The  first 
harmonics  are  called  fundamentals.  Theoretically  an  infinite 
number  of  terms  is  required  to  represent  most  non-sinusoidal 
voltages  or  currents,  but  for  most  practical  purposes  the  first 
few  terms  are  sufficient.  It  is  seldom  necessary  to  go  beyond 
the  term  of  eleventh  frequency.  For  most  wave  forms  the  series 
converges  rapidly. 

Under  ordinary  conditions,  the  first  harmonic,  or  fundamental 
as  it  is  called,  (this  includes  both  the  sine  and  cosine  term  of 
fundamental  frequency)  is  the  most  important  term  in  the  series 
and  it  has  by  far  the  largest  amplitude.  Any  harmonic  may  be 
absent  from  a  voltage  or  current  wave.  On  the  other  hand, 
certain  conditions  may  very  much  exaggerate  a  harmonic  in  a 
current  wave.  Also,  certain  harmonics  may  be  present  in  a 
current  wave  which  are  not  present  in  the  voltage  causing  that 
wave.  This  always  occurs  when  a  voltage  is  impressed  on  a 
circuit  whose  resistance,  inductance  or  capacitance  is  a  function 
of  the  current. 

The  sine  and  cosine  terms  of  the  Fourier  series  may  be  com- 
bined to  give  a  series  involving  either  sine  or  cosine  terms  alone. 
For  example,  equation  (2)  may  be  written 

e  =  A0  +  Ci  sin  (co£  +  Oi)  +  C2  sin  (2o>*  +  02) 

+  C3  sin  (3orf  +  03)  +  etc.     (3) 
or 
e  =  Ao  +  Ci  cos  (cot  -  0/)  +  C2  cos  (2orf  -  02') 

+  C3  cos  (3wt  -  03')  +  etc.    (4) 


The  angles  0  in  equation  (3)  are  the  angles  of  lead  between  the 
harmonics  of  the  sine  series  and  the  corresponding  sine  com- 
ponents in  equation  (2).  The  angles  0'  in  equation  (4)  are  the 
angles  of  lag  between  the  harmonics  of  the  cosine  series  and  the 
corresponding  cosine  components  in  equation  (2).  All  the  angles 
0  and  6'  in  equations  (3)  and  (4)  are  measured  on  the  scales  of 

angles  for  the  harmonics.     A  phase  displacement  of  0  degrees  for 

/\ 

the  nth  harmonic  corresponds  to  a  displacement  of  -  degrees  for 

the  fundamental.  See  Fig.  22.  If  the  phase  angles  for  the  har- 
monics are  measured  on  the  scale  of  angles  for  the  fundamental, 
equations  (3)  and  (4)  become 


74 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


i  sin  (ut  +  «i)  +  C2  sin  2(co£  +  .«2) 

+  C3  sin  3(co/  +  as)  +  etc.     (5) 
+  Ci  cos  (to*  -  a/)  +  C2  cos  2(co£  -  «2') 

+  C3  cos  3M  -  03')  +  etc.     (6) 


where 


=  0^3  and 


«3r  =  0  03;.     In  general  an  =  -  Bn  and  «„'  =  -  0n'  where  n  is  the 

O  7i  7i 

order  of  the  harmonic  considered. 

Since  sine  and  cosine  functions  of  time  differ  in  phase  by  90 
degrees,  the  cosine  function  leading,  the  A  and  B  terms  in 
equation  (2)  are  in  quadrature.  Therefore 


VAS  + 

Bi2         tan  0i  =  -j- 
R« 

tan  02'   =  1-r 
tan  0n'   =  4" 

£>n 

In  general 

AJ  2 

Bo2!        tan  02  =  -; 

A2 

5 

,    =    VAn2   + 

An 

7'       \»^  Resultant 


\    ,x  Fundamental 


FIG.  22. 


Since,  ordinarily,  the  current  and  voltage  waves  of  alternators 
operating  under  steady  conditions  are  symmetrical  with  respect 
to  the  axis  of  time,  the  constant  A0  in  the  Fourier  series  for  a 
current  or  a  voltage  may  usually  be  omitted. 


NON-SINUSOIDAL  WAVES  75 

A  wave  containing  a  fundamental  and  third  and  seventh 
harmonics  is  shown  in  Fig.  22.  The  fundamental  and  the  har- 
monics are  shown  by  full  lines.  The  resultant  wave  is  shown 
dotted. 

Effect  of  Even  Harmonics  on  Wave  Form. — Symmetrical 
waves,  i.e.,  waves  which  have  exactly  similar  positive  and  nega- 
tive loops,  cannot  contain  even  harmonics,  since  the  phase  of 
any  even  harmonic  with  respect  to  the  fundamental  will  be 
opposite  in  the  two  halves  of  the  wave.  Let  the  following 
Fourier  series  represent  a  voltage  which  contains  both  odd  and 
even  harmonics: 
e  =  Ei  sin  (o>*  +  ft)  +  #2  sin  (2o><  +  ft) 

+  #3  sin  (3o>*  +  ft)  +  #4  sin  (4o>*  +  ft) 

+  Eb  sin  (5co*  +  0s)  +  etc.  (7) 

If  this  wave  is  symmetrical  with  respect  to  its  positive  and 
negative  loops,  its  instantaneous  values  at  two  instants  of  time, 

/  T^  /  "1  \ 

such  as  t  and  (t  +  2" )  =  ( t  +  s> )  ,  which  are  separated  by  one 

half  period,  must  be  equal  in  magnitude  but  opposite  in  sign. 
Equation  (7)  gives  the  instantaneous  value  of  e  for  a  time  t  =  t. 

For  t  =  (  ^  +  of )  the  instantaneous  value  of  e  is 
e'  =  E,  sin  (««+•»  +  0i)  +  E2  sin  fat  +  ~  +  ft) 

+  E3  sin  fat  +  j|?  +  03)  +  E,  sin  fat  +  y  +   04) 
+  #5  sin  fat  +  I*  +  ft)  +  etc.    (8) 

Remembering  that  o>  =  2?r/  and  also  that  a  phase  displacement 
of  any  whole  number  of  wave  lengths,  i.e.,  any  whole  number  of 
2w  radians,  is  equivalent  to  zero  displacement  so  far  as  phase 
relations  are  concerned,  equation  (8)  may  be  reduced  to  the 
following  form: 
e'  =  #1  sin  (wt  +  TT  +  ft)  +  #2  sin  (2orf  +  ft) 

+  #3  sin  (3o>J  +  TT  +  ft)  +  #4  -sin  (4o>*  +  04) 

+  #5  sin  (5co2  +  7i •  +  0s) 
=  -#i  sin  (w$  +  ft)  +  #2  sin  (2orf  +  02) 

-#3  sin  (3o><  +  0,)  +  #4  sin  (4orf  +  ft) 

-#5  sin  (5orf  +  ft)  +  etc.  (9) 


76  PRINCIPLES  OF  ALTERNATING  CURRENTS 

It  will  be  seen,  by  comparing  equations  (7)  and  (9),  that  while 
the  fundamentals  and  the  corresponding  odd  harmonics  for 
points  one  half  a  period  apart  are  opposite  in  phase,  the  even 
harmonics  are  in  phase.  Therefore  the  two  halves  of  a  wave 
containing  even  harmonics  cannot  be  alike  in  shape. 

That  the  two  halves  of  a  wave  which  contains  even  harmonics 
are  not  alike,  i.e.,  that  a  wave  containing  even  harmonics  is  not 

symmetrical,  is  also  shown  by 
Fig.  23,  which  shows  a  wave 
containing  a  fundamental  and 
a  second  harmonic. 

Since  the  voltage  waves  of 
commercial     alternators    are 
symmetrical,  they  cannot  con- 
tain even   harmonics.     Even 
FlG  23  harmonics  may,  however,  oc- 

cur in  a  current   wave  even 

though  they  are  not  present  in  the  voltage  producing  it,  but  the 
conditions  under  which  they  occur  seldom  arise  in  practice. 
Even  harmonics  will  occur  in  the  current  wave  when  an  alternat- 
ing voltage  is  impressed  on  the  winding  of  an  inductive  circuit 
having  an  iron  core  which  is  magnetized  in  a  fixed  direction  by 
any  means,  such  as  a  constant  direct  current  in  a  second  winding 
surrounding  the  core.  In  this  case  the  self-induction  of  the 
winding  carrying  the  alternating  current  is  a  function  of  the 
current  and  will  be  different  for  positive  and  negative  values 
of  the  alternating  current.  This  should  be  better  understood 
after  studying  Chapter  V. 

Waves  which  have  the  Halves  of  the  Positive  and  of  the 
Negative  Loop  Symmetrical. — Waves  which  are  symmetrical 
with  respect  to  their  positive  and  negative  loops  cannot  contain 
even  harmonics.  If  the  halves  of  the  positive  and  of  the  nega- 
tive loop  are  also  to  be  symmetrical,  the  fundamental  and  all 
harmonics  must  pass  through  zero  at  the  same  instant.  For  this 
condition  to  be  fulfilled,  either  the  sine  or  the  cosine  terms  in  the 
Fourier  series  given  in  equation  (2)  must  be  zero.  All  even 
harmonics  and  the  constant  AQ  must  also  be  zero  if  the  positive 
and  negative  loops  are  to  be  symmetrical. 

If  the  equation  for  the  wave  is  written  in  sine  or  cosine  terms 


NON-SINUSOIDAL  WAVES  77 

only  (See  equations  (3)  and  (4),  page  73),  the  phase  angles  of  all 
harmonics  must  be  either  zero  or  180  degrees  when  the  phase  angle 
for  the  fundamental  is  made  zero  by  considering  time,  t,  zero 
when  the  fundamental  is  zero. 

Changing  the  Reference  Point  from  which  Angles  and  Time 
are  Measured  in  a  Complex  Wave. — It  is  frequently  convenient 
and  often  necessary,  when  considering  complex  waves,  to  change 
the  position  point  from  which  angles  and  time  are  reckoned. 
One  case  in  which  this  is  necessary  is  when  the  wave  forms  of 
two  waves,  whose  Fourier  equations  are  known,  are  to  be 
compared. 

For  two  waves  to  be  similar,  they  must  not  only  contain  like 
harmonics,  but  the  relative  magnitudes  of  the  harmonics  and 
fundamental  and  their  phase  relations  must  be  alike  in  the  two 
waves. 

A  glance  at  the  Fourier  equations  of  any  two  waves,  which  are 
to  be  compared  for  wave  form,  will  tell  whether  they  contain  the 
same  harmonics.  The  relative  magnitudes  of  the  harmonics 
and  the  fundamental  may  be  easily  determined  from  the  coef- 
ficients of  the  Fourier  series,  but  unless  the  points  from  which 
angles  are  measured  occupy  the  same  positions  with  respect  to 
the  fundamentals,  the  relative  phase  of  the  harmonics  in  the 
waves  will  not  be  obvious. 

If  the  points  from  which  angles  are  reckoned  do  not  occupy  the 
same  relative  positions  with  respect  to  the  fundamentals,  it  will 
be  necessary  to  shift  one  of  them  until  they  do. 

Consider  the  following  two  waves: 

el  =  100  sin  (orf  +  30°)  +  40  sin  (3ut  +  75°) 

+  20  sin  (but  -  60°)  (10) 

62  =  150  sin  M  -  10°)  +  60  sin  (3coZ  -  45°) 

-  30sin(5coJ  -  80°)  (11) 

The  waves  contain  like  harmonics.  The  magnitudes  of  the 
fundamental  and  harmonics  in  the  second  wave  are  fifty  per  cent, 
greater  than  in  the  first  wave.  The  relative  magnitudes  of  the 
harmonics  and  fundamental  are  the  same  in  both  waves. 

To  compare  the  phase  relations  of  the  harmonics  and  funda- 
mentals, one  wave  must  be  shifted  in  phase  by  an  amount  equal 
to  the  difference  between  the  fundamental  phase  angles  of  the 


78  PRINCIPLES  OF  ALTERNATING  CURRENTS 

two  waves.  This  can  be  accomplished  either  by  shifting  the  first 
wave  minus  40  degrees  or  the  second  plus  40  degrees.  Shift  the 
second  wave.  Since  any  phase  displacement  of  a  degrees  for 
the  fundamental  corresponds  to  a  phase  displacement  of  no.  de- 
grees for  the  nth  harmonic,  it  follows  that  if  40  degrees  be 
added  to  the  fundamental  phase  angle,  3  X  40  =  120  degrees 
must  be  added  to  the  phase  angle  of  the  third  harmonic  and 
5  X  40  =  200  degrees  must  be  added  to  the  phase  angle  of  the 
fifth  harmonic.  Shifting  the  fundamental  of  the  second  wave 
plus  40  degrees  in  phase  gives 

e2'  =  150  sin  («*  +  30°)  +  60  sin  (3co£  +  75°) 

-  30  sin  (5co£  +  120°) 

=  150  sin  (cot  +  30°)  -f  60  sin  (3urf  +  75°) 

+  30  sin  (5ut  -  60°)  (12) 

From  equations  (10)  and  (12)  it  is  obvious  that  the  differences 
in  phase  between  the  harmonics  and  fundamental  are  alike  in  the 
two  waves. 

Fourier  Series  for  Rectangular  and  Triangular  Waves. — The 
Fourier  equations  for  a  rectangular  and  for  a  triangular  wave  are 
interesting  since  such  waves  represent  extreme  cases  of  a  flattened 
and  a  peaked  sine  wave.  Neither  of  these  wave  forms  could  be 
exactly  attained  in  practice  nor  would  either  be  desirable  even  if  it 
could  be  secured. 

The  Fourier  series  for  rectangular  and  for  triangular  voltage 
waves  are  given  by  equations  (13)  and  (14)  respectively. 

e  =  El  sin  cot  +  -E±  sin  3u>£  +  KE1  sin  fat 

o  O 

+  7  El  sin  7 (at  +  etc.       (13) 
e  =  EI  sin  cot  —  —  El  sin  3co£  +  ^  EI  sin  5«£ 

-  ^  #1  sin  7co£  +  etc.     (14) 

Both  the  rectangular  and  the  triangular  wave  have  similar 
positive  and  negative  loops  and  therefore  contain  only  odd 
harmonics.  Also  since  the  halves  of  each  loop  are  similar,  all  the 
phase  angles  are  zero. 


NON-SINUSOIDAL  WAVES 


79 


The  manner  in  which  the  wave  form  given  by  equation  (13) 
approaches  that  of  a  rectangular  wave,  as  successive  terms  are 
added,  is  illustrated  in  Fig.  24. 


-- 


0 

/<             >\ 

f 

,'                            TT 

e  -  J57j  sin  w«+i  Ev  sin  Swt 

1 

X 


FIG.  24. 


Although  an  infinite  number  of  terms  would  be  required  to 
exactly  represent  a  rectangular  wave,  a  fairly  good  approximation 
to  such  a  wave  is  obtained  with  a  comparatively  few  terms. 
Even  with  the  four  terms  plotted  in  Fig.  24,  the  resultant  is 
rapidly  flattening  out  and  approaching  the  rectangular  form. 
Ordinary  waves  met  in  practice  can  usually  be  represented  with 
sufficient  accuracy  for  most  work  by  a  comparatively  few  terms 
of  their  Fourier  equations. 

Measurement  of  Current,  Voltage  and  Power  when  the  Wave 
Form  is  not  Sinusoidal.  —  Since  the  average  torque  producing  the 
deflection  in  the  electrodynamometer  and  iron-vane  types  of 
ammeter  is  proportional  to  the  average  square  of  the  current 
in  their  coils,  such  instruments  may  be  used  to  measure  current 
when  the  wave  form  is  not  sinusoidal.  This  statement  is  not 
strictly  correct  for  the  iron-vane  type  of  instrument,  since  the 
eddy  currents  and  hysteresis  in  the  iron  vane  may  seriously 
affect  the  readings  when  pronounced  high-frequency  harmonics 
are  present.  However,  when  used  on  circuits  of  commercial 
frequencies  and  wave  forms,  the  readings  are  approximately 
correct.  Voltmeters  are  usually  of  the  electrodynamometer 


80  PRINCIPLES  OF  ALTERNATING  CURRENTS 

type.  Iron  vanes  are  seldom  used  except  in  the  cheaper  instru- 
ments. For  an  electrodynamometer  type  of  voltmeter  to  indi- 
cate correctly,  the  current  through  it  at  each  instant  must  be 
strictly  proportional  to  the  instantaneous  voltage  across  its 
terminals. 

The  general  effect  of  inductance  in  a  circuit  is  to  damp  out 
the  harmonics  in  the  current  to  a  greater  and  greater  degree  as 
their  frequency  increases.  (See  Chap.  VII,  page  208.)  There- 
fore, since  the  coils  of  an  instrument  cannot  be  made  without 
inductance,  the  current  in  a  voltmeter  cannot  be  strictly  propor- 
tional, at  each  instant,  to  the  voltage  across  its  terminals,  except 
when  the  voltage  is  sinusoidal.  If  the  non-inductive  resistance 
in  series  with  a  voltmeter  is  large  compared  with  the  inductance 
of  its  coils,  the  current  through  it  will  be  substantially  propor- 
tional, at  each  instant,  to  the  voltage  across  its  terminals,  pro- 
vided the  voltage  does  not  contain  very  pronounced  harmonics 
of  high  frequency.  Both  the  electrodynamometer  and  iron-vane 
types  of  voltmeter,  when  used  on  circuits  of  commercial  fre- 
quencies and  wave  forms,  indicate  root-mean-square  or  effective 
voltage  to  a  high  degree  of  precision. 

For  very  high  frequencies  the  hot-wire  type  of  ammeter  and 
voltmeter  must  be  used.  These  instruments  depend  on  the  ex- 
pansion of  a  fine  wire  due  to  the  heating  caused  by  the  current 
in  the  wire.  The  effect  of  the  change  in  length  of  the  wire  is 
multiplied  and  causes  a  pointer  to  move  over  a  graduated  scale. 
To  measure  current  the  wire  is  placed  in  series  with  the  circuit 
carrying  the  current  to  be  measured.  To  measure  voltage  the 
wire  is  shunted  across  the  circuit  in  series  with  a  non-inductive 
resistance.  Since  the  strength  of  a  current  is  defined  in  terms 
of  its  heating  effect,  the  hot-wire  type  of  instrument  will  indicate 
the  average  square  value  of  the  current  in  the  hot  wire.  The 
inductance  of  such  an  instrument  is  extremely  small,  being  due  to 
the  inductance  of  a  straight,  fine  wire  only  a  few  inches  in  length. 
For  this  reason  hot-wire  instruments  may  be  used  on  very  high 
frequency  circuits.  They  do  not  hold  their  calibration  well  and 
are  therefore  not  satisfactory  for  general  use  on  circuits  of 
commercial  frequency. 

The  electrodynamometer  type  of  wattmeter  will  indicate  true 
average  power  provided  the  current  in  its  current  coil  is  equal 


NON-SINUSOIDAL  WAVES  81 

or  proportional,  at  each  instant,  to  the  current  in  the  circuit, 
and  the  current  in  its  potential  circuit  is  proportional,  at  each 
instant,  to  the  voltage  across  its  terminals.  The  instrument 
is  subject  to  the  same  limitations  as  the  voltmeter,  but  by  proper 
design  it  may  be  made  to  indicate  true  average  power  when  used 
in  circuits  of  ordinary  commercial  frequency  and  wave  form. 

Determination  of  Wave  Form. — The  usual  way  of  determining 
the  wave  form  of  a  current  or  voltage  is  by  means  of  an  oscillo- 
graph. The  oscillograph,  in  its  simplest  form,  consists  of  a 
vibrating  element  made  of  a  single  loop  of  very  fine  wire  which 
is  stretched  over  two  bridges.  The  straight,  parallel  sides  of  the 
loop  between  the  bridges  are  from  0.2  to  0.25  millimeter  apart, 
and  lie  in  a  strong,  uniform  field  between  the  poles  of  an  electro- 
magnet which  is  excited  with  direct  current.  The  plane  of  the 
loop  is  parallel  to  the  axis  of  the  magnetic  field.  A  very  small, 
light  mirror  is  attached  to  the  parallel  sides  of  the  loop  midway 
between  the  bridges. 

When  a  current  is  passed  through  the  loop  it  will  flow  down 
one  side  and  up  the  other,  causing  the  two  sides  of  the  loop  to 
deflect  in  opposite  directions.  The  mirror  will  tilt  through  an 
angle  which  is  proportional  to  the  strength  of  the  current.  If  a 
spot  of  light  from  a  source  which  approximates  a  point  is  re- 
flected on  a  screen  by  the  mirror,  the  spot  will  move,  when  the 
mirror  is  deflected,  through  a  distance  which  is  proportional  to 
angular  displacement  of  the  mirror.  This  assumes  that  the 
displacement  of  the  mirror  is  small.  The  loop  with  its  mirror  is 
immersed  in  oil  to  damp  its  vibrations  and  make  it  dead  beat. 
When  the  wave  form  is  to  be  photographed,  the  spot  of  light 
is  reflected  on  a  revolving  drum  which  carries  the  film.  The 
rotation  of  this  drum  gives  the  time  element,  i.e.,  the  abscissa 
of  the  wave,  and  the  displacement  of  the  spot,  which  must  be 
parallel  to  the  axis  of  the  drum,  gives  the  other  element,  i.e.,  the 
ordinate  of  the  wave.  To  prevent  overlapping  of  the  waves  on 
the  drum,  a  shutter  is  interposed  between  the  light  source  and  the 
drum  and  remains  open  only  during  one  revolution  of  the 
drum.  When  the  wave  is  to  be  observed  or  traced  on  a  screen, 
the  time  element  of  the  wave  is  obtained  by  placing  a  revolving 
mirror  between  the  source  of  light  and  the  screen.  To  keep  the 
image  of  the  wave  fixed  on  the  screen  this  mirror  must  revolve 
synchronously  with  the  frequency  of  the  circuit. 


82  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  dimensions  of  the  vibrating  system  must  be  such  that  its 
natural  frequency  of  vibration  shall  be  at  least  fifty  times  as 
great  as  the  frequency  of  the  highest  harmonic  to  be  detected. 
The  minimum  free  period  which  it  is  practical  to  obtain  is  about 
0.0001  second,  but  for  most  commercial  work  a  period  of  0.0002 
second  is  sufficiently  low. 

For  current  measurements  the  vibrating  element  is  shunted 
across  the  terminals  of  a  non-inductive  shunt  which  is  placed  in 
series  with  the  circuit  carrying  the  current  whose  wave  form  is  to 
be  determined.  Since  the  vibrating  element  is  sensibly  non- 
inductive,  the  current  it  carries  at  each  instant  will  be  directly 
proportional  to  the  current  in  the  circuit.  For  voltage  measure- 
ment the  vibrating  element  is  connected  in  series  with  a  suitable 
non-inductive  resistance  and  is  then  shunted  across  the  circuit 
whose  voltage  wave  form  is  to  be  determined. 

Effective  Value  of  a  Non-sinusoidal  Electromotive  Force  or 
Current. — Let 

e  =  Enl  sin  (coZ  +  00  +  Em,  sin  (3co*  +  03) 

+  Emb  sin  (5w<  +  05)  +  etc.  (15) 

be  an  alternating  electromotive  force  of  periodic  time  T  =  — » 

CO 

containing  a  fundamental,  Em\  sin  (at  -f  61),  and  odd  harmonics, 
Em3  sin  (3co£  +  63),  Emb  sin  (5wt  +  05),  etc.,  whose  periods  are  odd 

multiples  of  the  fundamental  period  — .     The  effective  or  root- 

co 

mean-square  value  of  the  electromotive  force  is  given  by 


T 

sin  M  +  00  +  Ems  sin  (3«J  +  03) 


.  }    dt     (16) 


This  involves  squared  terms  of  the  general  form 


and  product  terms  of  the  general  form 
T 


1  CT 

^         [Emk  sin  (kwt  +  0,)  X  EmQ  sin  (qoit  +  0,)  }dt 

1  Jo 


NON-SINUSOIDAL  WAVES  83 

Each  of  the  squared  terms,  on  integration  over  a  complete 
cycle,  becomes  equal  to  one-half  the  square  of  its  maximum 
value,  since  the  average  product  of  two  sine  terms  of  like  fre- 
quency is  one-half  the  product  of  the  maximum  values.  Each  of 
the  product  terms  of  unlike  frequency  becomes  zero  on  integra- 
tion over  a  complete  cycle,  since  the  average  value  of  the  product 
of  two  sine  terms  of  unlike  frequency  is  zero. 

Consider  the  product  of  two  sine  terms  of  like  frequency. 
Let  o>£  =  a. 

f  (a)  =  A  sin  (a  +  0)  X  B  sin  (a  +  0') 

=  AB  (sin  a  cos  0  +  cos  a  sin  0)  (sin  a  cos  B'  +  cos  a  sin  B') 

=  AB:  sin2  a  cos  0  cos  0'  -f  sin  a  cos  a  cos  0  sin  B' 

4-  cos  a  sin  a  sin  0  cos  B'  +  cos2  a  sin  0  sin  B'  | 

But 

1  —  cos  2a  14-  cos  2a      .  sin  2a 

sin2  a  =  -  — ,    cos2  a  =  -  — ,    Sin  a  cos  a  =  — » — . 

Z  Z  Z 

Therefore 
f(a)  =  AJ5J-        C°S    -  cos  0  cos  0'  +  ^^o^"  cos  ^  s*n  ^ 

.   sin  2a    ,  14  cos  2a    . 

H —  sin  0  cos  r  H —    — = —    -  sin  0  sin 

Z  Z 

and 

1   r2*"  f  1  1  1 

f  (o)do  =  AB  \  0  cos  0  cos  0'  4  0  4  0  +  =  sin  0  sin  0'  [ 

27T  Jo  1 2  Z 

=  -~  cos  (0-0')  (17) 

Now  consider  the  product  of  two  sine  terms  of  unlike  frequency, 
f  (a)  =  A  sin  (a  +  0)  X  B  sin  (no  +  0n) 

=  AB{  (sin  a  cos  0  4-  cos  a  sin  0)  (sin  na  cos  0n 

4-  cos  na  sin  0n)j 

=  AB{sin  a  sin  na  cos  0  cos  0n  4-  sin  a  cos  wa  cos  0  sin  0W 
4-  cos  a  sin  na  sin  0  cos  Bn  4-  cos  a  cos  na  sin  0  sin  0n  J 
where  n  is  any  integer  which  is  greater  than  unity. 

Since  0  and  0n  are  constants 
f(a)  =  AB{Ki  sin  a  sin  na  4  ^2  sin  a  cos  na 

4-  -K3  cos  a  sin  na  4  K4  cos  a  cos  na} 


84  PRINCIPLES  OF  ALTERNATING  CURRENTS 

But 

sin  x  sin  y  =  ~  j  cos  (x  —  y)  —  cos  (x  +  y) 


cos  x  sin  y 


L'si 

sin  (x  +  y)  —  sin  (x  —  y) 


sin  x  cos  y  =  ~  { sin  C&  +  2/)  +  sin  (x  —  ?/) 
1 


If 


cos  x  cos  y  =  s  j  cos  (a;  +  y)  +  cos  (a;  —  y) 
Therefore 

f (a)  =  AB  -^  [cos  (1  —  n)a  —  cos  (1  +  n)a] 

2 

rr 

+  — 2  [sin  (1  +  ri)a  +  sin  (1  —  w)a] 


+  -~  [sin  (1  +  ri)a  —  sin  (1  —  n)a] 

+  -?r[cos  (1  +  n)a  +  cos  (1  —  ri)a] 


and 


=  0 


(18) 


The  root-mean-square  or  effective  value  of  a  non-sinusoidal 
voltage  is  therefore 


E  = 


etc. 


+ 


+  Ei*  +  etc. 


(19) 
(20) 


where  the  E's  with  the  subscript  m  are  maximum  values  of  the 
components  of  the  voltage.  Without  the  subscript  m  they  are 
the  effective  values. 

Similarly,  if 
i  =  Imi  sin  (co£  +  0i  -f  ai)  +  Ima  sin  (3coi  +  03  +  0:3) 

-f-  7W5  sin  (5oo£  -j-  ^5  -f-  #5)  4~  etc. 

is  an  alternating  current  of  periodic  time  T  =  — ,  its  root-mean- 

co 

square  or  effective  value  is  given  by 
7  = 


+  /s2  +  /52  +  etc. 


(21) 


NON-SINUSOIDAL  WAVES  85 

where  the  /'s  with  the  subscript  m  are  maximum  values  of  the 
components  of  the  current.  Without  the  subscript  m  they  are 
the  effective  values. 

Power  when  the  Electromotive  Force  and  Current  are  Non- 
sinusoidal  Waves.  —  Let 
e  =  Eml  sin  (<*t  +  00  +  Em3  sin  (3orf  +  0») 

+  Ems  sin  (5orf  +  05)  +  etc. 
and 
i  =  Imi  sin  M  +  0/)  +  /„,  sin  (3orf  +  *.') 

+  7m5  sin  (5orf  -f  06')  +  etc. 
represent  an  electromotive  force  and  current  respectively,  whose 

periodic  time  is  T  =  —  .     The  average  power  is 


P  =  ~ 


CO 
T 


sm  (cat  +  0i)  +  EmZ  sin 


+  Em5  sin  (5co2  +  05)  +  etc.}  X  {/mi  sin  (cot  +  0/) 

+  /ma  sin  (3co£  +  03')  +  /mo  sin  (5co^  -f-  0s')  +  etc.}dt 
This  involves  product  terms  of  like  and  unlike  frequency  of 
the  forms 


and 


1  CT 

^       {Emk  sin  (kwl  +  0*)  X  /m*  sin  (kwt  + 
^  Jo 

1  fr 

^       {Emk  sin  (fc««  +  0A)  X  Imq  sin  (g««  +  0/ 
-/Jo 


On  integration,  the  product  terms  of  like  frequency  become 
(see  equation  (17),  page  83) 

^^  cos  (0,  -  0*') 

while  the  product  terms  of  unlike  frequency  become  zero  on 
integration.     (See  equation  (18),  page  84) 

The  average  power  in  a  circuit  having  harmonics  in  both  cur- 
rent and  voltage  is  therefore 
p  =  E  cog    $i  _  ,i/    +  E  cog  (^  _  03/) 


+     mbmb  cos  (05  -  05')  +  etc.        (22) 

P  =  E1I1   cos    (0i  -  0/)  +  #3/3   cos    (03  -  03') 

+  #5/5  cos  (05  -  05')  +  etc.  (23) 


86  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  letters  E  and  /  in  equations  (22)  and  (23)  with  the  sub- 
script m  represent  maximum  values.  Without  the  subscript  m 
they  represent  root-mean-square  values. 

If  a  harmonic  occurs  in  the  current  and  is  not  present  in  the 
voltage,  or  vice  versa  if  a  harmonic  occurs  in  the  voltage  and 
is  not  present  in  the  current,  it  will  contribute  nothing  to  the 
average  power  developed. 

Power  -factor  when  the  Current  and  Voltage  are  not  Sinu- 
soidal. —  The  power-factor  of  a  circuit  is  denned  as  the  ratio 
of  the  true  average  power  to  the  volt-amperes.  This  definition 
is  independent  of  wave  form. 

f  True  power         P 

Power-factor  =  ^^-v  -  =TTT  (24) 

Volt-amperes      E  I 

#!/!  cos  (0i  -  0i')  +  #3/3  cos  (03  -  03')  +  etc. 


Although  a  harmonic  which  occurs  in  the  current  of  a  circuit 
and  does  not  occur  in  the  voltage  does  not  contribute  to  the 
average  power  it  does  increase  the  root-mean-square  or  effective 
value  of  the  current  required  to  produce  the  power.  Therefore 
the  power-factor  of  a  circuit  containing  a  harmonic  in  its  current 
which  is  not  present  in  its  voltage  cannot  be  unity.  Similarly, 
the  power-factor  of  a  circuit  containing  a  harmonic  in  its  voltage 
which  is  not  present  in  its  current  cannot  be  unity. 

The  only  way  the  power-factor  can  be  unity  is  for 

cos  (0i  -  0i')  =  cos  (0S  -  ezf)  =  etc.  =  1 

and  #1      E8 

j-=-j-    =  etc. 
Ii       is 

When  the  current  and  voltage  waves  are  of  different  form,  the 
maximum  power-factor,  for  fixed  effective  values  of  current  and 
voltage,  will  obviously  occur  when  the  phase  displacement 
between  the  current  and  voltage  is  that  which  makes  the  power 
a  maximum. 

From  equation  (25)  it  is  obvious  that  unity  power-factor  can 
occur  only  when  the  current  and  voltage  waves  are  exactly 
similar  in  form  and  have  no  phase  displacement  with  respect  to 
each  other.  In  other  words,  both  waves  must  contain  like 
harmonics  and  these  harmonics  must  have  the  same  relative 
magnitudes  and  the  same  relative  phase  relations.  If  there  is  a 


NON-SINUSOIDAL  WAVES  87 

harmonic  in  the  current  which  is  not  present  in  the  voltage,  the 
power-factor  cannot  be  unity,  since  this  harmonic  will  contribute 
to  the  root-mean-square  value  of  the  current  without  adding  to 
the  power  developed.  A  similar  statement  is  of  course  true 
regarding  a  harmonic  in  the  voltage. 

There  are  many  cases  where  harmonics  are  present  in  the 
current  and  are  not  present  in  the  voltage  causing  it.  Such  a 
condition  always  occurs  when  a  voltage  is  impressed  on  a  circuit 
whose  inductance  is  a  function  of  the  current.  Since  the  induc- 
tance of  a  circuit  is  defined  as  the  flux  linkages  per  unit  current, 

i.e.,  as  L  =  N  -JT,  the  inductance  of  all  circuits  containing  iron 

must  be  a  function  of  the  current. 

Since  the  power-factor  of  a  circuit  is  equal  to  the  ratio  of  true 
power  to  volt-amperes,  power-factor  might  be  defined  as  the 
ratio  of  the  actual  power  to  the  maximum  power  that  could  be 
obtained  with  the  given  current  and  voltage. 

Example  of  the  Calculation  of  Effective  Values  of  Current  and 
Voltage,  Average  Power  and  Power-factor  for  a  Circuit  when 
the  Fourier  Equations  of  its  Current  and  Voltage  are  Known. — 
When  the  voltage 

v  =  200  sin  (377*  +  10°)  +  75  sin  (1131*  +  30°) 

+  50  sin  (1885*  +  50°) 
is  impressed  on  a  certain  circuit,  the  current  is 

i  =  8.51  sin  (377*  +  11°18)  +  8.04  sin  (1131*  +  74°33) 

+  7.50  sin  (1885*  +  84°32) 

v         /(200)2  +  (75)2  +  (50)2 

\  2 

=  155.1  volts. 


(8.51)2  +  (8.04)2  +  (7.50)' 


=  9.84  amperes. 
P= 


2 

,   75  X  8.( 

J.  A  .  XO  ) 

M 
ros  (30°        74°W 

2 

so  ; 

*  7'5°   nn«  f*n°         $U°29^ 

851  +  216  +  155  =  1222  watts. 

1222 
Power-factor  -  ^  =  0.801 


88 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Effective  Value  of  an  Alternating  Electromotive  Force  or 
Current  from  a  Polar  Plot. — If  an  electromotive  force  or  current 
wave  has  been  analysed  and  the  components  of  its  Fourier 
series  have  been  determined  in  magnitude,  its  effective  value 
may  be  readily  calculated.  Frequently,  however,  the  time  re- 
quired in  such  an  analysis  is  excessive  for  the  purpose  in  hand. 
In  such  a  case  the  following  graphical  method  gives  satisfactory 
results. 

Let  Fig.  25-a  represent  an  alternating-current  wave  which  has 
been  determined  by  means  of  an  oscillograph.  Divide  either 

half  of  the  wave  into  any 
convenient  number  of 
equal  parts,  the  number 
depending  on  the  preci- 
sion desired,  and  erect 
ordinates  at  the  points 
of  division.  These  ordi- 
nates represent  instan- 
taneous values,  i,  of  the 
current. 

Make  a  polar  plot  of  the  wave  as  shown  in  Fig.  25-&,  in  which 
any  radius  such  as  that  at  e  is  equal  in  length  to  the  ordinate  at  e 
in  the  left-hand  half  of  the  figure  and  is  laid  off  at  an  angle  from 

CLC 

the  base  line  equal  to  180  X  —  degrees. 

An  element  of  area  of  the  polar  plot  enclosed  between  two 
radii  differing  in  direction  by  an  angle  d0  is  a  triangle  of  base 

idQ,  altitude  i  and  area  ^  d6.     The  area  of  the  polar  plot  is 
therefore 


abed 
(a) 


e    f  g 


FIG.  25. 


where    /    is   the   effective   or   root-mean-square   value    of   the 
current  and  T  is  the  time  of  a  complete  period. 

Planimeter  the  polar  diagram  and  determine  its  area.     Then 


Area  = 


(26) 


NON-SINUSOIDAL  WAVES  89 

The  effective  value  of  the  current  is  then  given  by 

/Area      _  -no     /-r— 
/  =      /—   -  =  0.798  V  Area 

\  I 

Analysis  of  a  Non -sinusoidal  Wave  and  Determination  of  Its 
Fundamental  and  Harmonics. — It  is  often  necessary  to  analyse 
a  voltage  or  current  wave,  obtained  by  an  oscillograph  or  other 
means,  into  the  components  of  its  Fourier  series  in  order  to 
determine  the  harmonics  present  and  their  magnitudes.  The 
Fourier  series  may  be  written, 

f  (a)  =  A  +  AI  sin  a  +  BI  cos  a  +  Az  sin  2a  +  J52  cos  2a  + 

.    .    .    +  An  sin  na  +  Bn  cos  na 

The  expression  for  the  constant  A  may  be  found  by  multi- 
plying the  equation  by  da  and  then  integrating  it  between  the 
limits  0  and  2ir.  For  example: 

r*  £2*  /»2ir  /»2ir 

f  (a)  da   =  A  I      da  +  AI  \      sin  ada  +  BI   I      cos  ada  + 
Jo  Jo  Jo 

.    .    .  +  An  I      sin  nada  +  Bn  I      cos  nada 
Jo  Jo 

All  terms  on  the  right-hand  side  of  the  equation,  except  the 
first,  reduce  to  zero  leaving 


i{a)da  =  2wA 

and 

1   T2ir 
A  -  ~         f(a)da 

**JO 

The  integral   I     f  (a)  da  is  the  net  area  enclosed  by  the  curve 

Jo 

represented  by  f(a)  for  a  complete  period  or  cycle.  Since  the 
positive  and  negative  loops  of  current  and  voltage  waves  of 
alternators  under  steady  conditions  are  symmetrical  with  respect 
to  the  axis  of  time,  the  net  area  represented  by  the  integral  and 
hence  the  constant  A  are  zero. 

To  obtain  the  second  constant,  AI,  multiply  both  sides  of  the 
equation  of  the  wave  by  sin  a  da  and  integrate  the  resulting 
equation  between  the  limits  0  and  2ir. 


90  PRINCIPLES  OF  ALTERNATING  CURRENTS 

rf*2ir 
f  (a)  sin  ada  =  A  I      sin  ada 
Jo 
pr   >  [2, 

+  AI  I      sin2  ada  +  B\   I      cos  a  sin  ada 
Jo  Jo 

r  pi- 

sin  2a  sin  ada  +  B2  I      cos  2a  sin  ada  +    • 
Jo 

/•27T  /»27T 

%  I     sin  na  sin  ada  +  Bn  I      i 

Jo  Jo 


•27T  /»27T 

+  An       sin  na  sin  ada  +  £n  I      cos  no.  sin  ada 


All  the  terms  on  the  right-hand  side  of  the  equation,  except 
the  second,  reduce  to  zero  leaving 

rf(a)  sin  ada  =  TT  AI 

1  C2" 
A]    =  -         f(a)  sin  ada 


The  constant  BI  may  be  found  in  a  similar  manner  by  multi- 
plying the  equation  of  the  wave  by  cos  ada  and  integrating 
between  the  limits  0  and  2?r.  In  general,  any  constant  may 
be  found  by  multiplying  both  sides  of  the  equation  by  da 
and  the  trigonometrical  function  which  appears  in  the  term 
containing  the  desired  constant,  and  then  integrating  the  result- 
ing equation  between  the  limits  0  and  27r.  For  example:  any 
constant  such  as  An  may  be  found  by  multiplying  both  sides 
of  the  equation  of  the  wave  form  by  sin  no.  da  and  then  integrating 
the  resulting  equation  between  the  limits  0  and  2ir.  Doing  this 
for  the  different  terms  gives 

i  r  2* 

AI  =  -  I      f(a)  sin  ada 
^Jo 

f  (a)  cos  ada 

1    T27r 

AZ  =  ~        f(o)  sin  2ada 
TJo 

1  f  27r 

B2  =  -        f  (a)  cos  2ada 
TJJ 


i 

An  =  -         f(a)  sin  nadfy 


f  (a)  cos  nada 


NON-SINUSOIDAL  WAVES 


91 


The  values  of  the  constants  may  be  found  by  evaluating  the 
integrals  by  means  of  a  graphical  method.  Let  the  wave  repre- 
sented by  f  (a)  =  y  be  plotted  as  shown  in  Fig.  26.  The  base  of 
this  figure  represents  a  complete  cycle  and  is  equal  to  2ir  radians. 


FIG.  26. 

Divide  the  figure  into  m  equal  parts  by  equally  spaced  ordi- 
nates.     The  distance  between  any  two  adjacent  ordinates  is 

equal  to  —  radians.     The  integral 

AI  =  -        f(a)  sin  ada 

""t/O 

may  then  be  written  as  a  summation. 

27T 

m 

where  yk  is  the  ordinate  of  the  curve  at  the  middle  of  the  kth 
space. 

A    =  -  1        in-  —  4          '    ?  %  4          •    5  ??   , 
in  <•  2  w  2m  2m 


.     /2m  -  1 
+  2/m  sm 


)S 


In  general 

2  (  /I  2r 


2r 


+  2/3  sin      n-    +    .     .     . 


/2m  -  1\    2ir 


\      ir 
)n- 


+  y,  cos 


>-)  +    .  +  V.  cos  (?=L- 


)»?? 
/    m 


92  PRINCIPLES  OF  ALTERNATING  CURRENTS 

If  the  equation  of  the  wave  is  desired  in  sine  or  cosine  terms 
alone,  the  constant  C  and  the  angles  0  or  0'  for  the  sine  or  cosine 
series  may  be  obtained  by  the  method  given  on  page  74.  For 
any  term  such  as  the  kth,  the  constant  C  and  the  angles  are 


., 

6k   =  tan   l  -p 

A-k 

Sk'  =  tan-'  £ 

-Ofc 

The  precision  of  the  constants  and  the  angles  is  increased 
by  making  m  large.  For  most  commercial  work  it  is  not  neces- 
sary to  carry  the  analysis  beyond  the  seventh  or  the  eleventh 
harmonic. 

Theoretically  it  is  easy  to  calculate  the  constants  by  the 
method  just  outlined.  The  real  difficulty  lies  in  the  amount  of 
time  involved.  This  may  be  reduced  by  arranging  the  work  in  a 
tabular  form.  The  forms  in  use  follow  a  method  introduced  by 
C.  Runge,*  in  which  the  calculation  of  the  constants  is  carried  out 
by  the  aid  of  a  systematically  arranged  schedule. 

Fischer-Hitmen  Method  of  Analysing  a  Periodic  Wave  into  the 
Components  of  Its  Fourier  Series.  —  TheFischer-Hinnenmethod,f 
which  is  also  known  as  the  "  Selected-ordinate  Method,"  reduces 
to  a  minimum  the  work  of  determining  the  constants  for  any 
definite  number  of  terms  of  the  Fourier  series  of  a  wave. 
It  is  based  on  two  mathematical  laws  which  can  easily  be 
demonstrated. 

Let  the  distance  between  any  two  points  a  and  b  corresponding 
to  a  whole  wave  length,  i.e.,  corresponding  to  360  degrees  or  2r 
radians,  be  divided  into  m  equal  parts.  Measure  the  length  of 
the  ordinate  at  the  beginning  of  each  part  for  any  harmonic 
such  as  the  kth.  Call  the  ordinates 

Ymi,  Ym2,  Fm3,  etc. 

where  the  first  subscript,  m,  indicates  the  number  of  parts 
into  which  the  wave  is  divided  and  the  second  subscript  indicates 
the  position  of  the  ordinate. 

*  Elektrotechnische  Zeitschrift,  1905,  p.  247. 

t  Elektrotechnische  Zeitschrift,  Vol.  22,  1901,  p.  296. 


NON-SINUSOIDAL  WAVES 

If  these  ordinates  are  added 

(360°  ) 

k-    -  -f  ka  > 
m  j 

+  Ak  sin 


93 


m 


+  Ak  sin  ^  3fc 


+ 


sin  ((m-  l)k  —  +  fca  1     (27) 
I  w  J 


The  angle  a  is  the  angular  distance,  measured  on  the  funda- 
mental scale  of  angles,  between  the  first  ordinate  and  the  point 
where  the  harmonic  considered  passes  through  zero  increasing  in  a 
positive  direction.  Ak  is  the  amplitude  of  the  harmonic.  (See 
Fig.  27.) 


FIG.  27. 


Expanding  the  angles  in  equation  (27)  gives 
27  Ym  =  Ak  sin  ka 


,  360°  ,  360°   .     , 

sin  k  --  cos  lea  +  Ak  cos  tc  -  sin  Ka 
m  m 


360° 

.k  sin  2fc  —    -  cos  ka  -f-  A*  cos  2k  -    -  sin  fca 
m  m 


.    ,          1W  360°         7 
sin  (m  —  I)  k  -    -  cos  fca 
m 

1M    360°      •        7 

H-  A*  cos  (m  —  l)k  -    -  sin  fca 
m 


94  PRINCIPLES  OF  ALTERNATING  CURRENTS 


/    •      /     Pi  ;  360°  o;  36° 

=  Ak  {  sm  Ka   1  +  cos  K  —    — r  cos  2k 

[  L  m  m 

1M    36°°1 

.    .    .  +  cos  (m  —  l)k 

/        r  •  1 36°°  .       .7  36°°  . 

-f-  Ak  {  cos  ka  sin  k  -    — \-  sm  2k  - 
{  L  m  m 

.    .  +  sin  (m  —  l)k 1  I      (28) 

m  J  } 

Since  the  sine  of  any  whole  number  times  360  degrees  is 
equal  to  zero  and  the  cosine  of  any  whole  number  times  360 
degrees  is  unity,  it  is  evident  from  inspection  of  equation  (28) 

k 

that  when  —  is  a  whole  number 
m 

2?  Ym  =  mAk  sin  ka  (29) 

The  sum  of  m  equally  spaced  ordinates  for  any  harmonic  such 
as  the  kth  is  therefore  equal  to  m  times  the  first  ordinate  when 
the  order,  k,  of  the  harmonic  is  any  whole  number  times  the 
number  of  ordinates,  m. 

k 

If  —  is  not  a  whole  number  for  the  harmonic  considered,  the 
m 

series  represented  by  equation  (28)  reduces  to  zero.     This  can  be 
shown  by  the  aid  of  the  two  following  trigonometrical  formulas. 

cos  6  +  cos  26  +  cos  30  +  . 


sin  6  +  sin  20  +  sin  30  +  ...... 

.     /m  -  l\  .        mB 
sin  (      o     j  °  sm  ~9 

+  sin  (m  -1)6=  -—-  (31) 

sin  - 

/b360°          k 
If  6  is  put  equal  to  -      -  and  —  is  not  a  whole  number,  equation 


(30)  reduces  to  —1  and  equation  (31)  reduces  to  zero.     There- 
is  not  a  whole  number 
Ym  =  Ak  {sin  ka  (1  -  1)  +  cos  fca(O)}   =  0        (32) 


k 

fore  when  —  is  not  a  whole  number 
m 


NON-SINUSOIDAL  WAVES 


95 


The  relations  given  in  equations  (29)  and  (32)  may  be  used 
to  determine  the  components  of  the  Fourier  series  for  any 
periodic  wave.  Let  Fig.  28  represent  a  fundamental  and  a  third 
harmonic  for  a  wave  containing  only  a  fundamental  and  third 
harmonic. 


FIG.  28. 

Erect  the  two  ordinates  a  and  b  360  degrees  apart,  taking  any 
convenient  point  such  as  a  for  the  origin.  Let  the  fundamental 
and  harmonic  of  the  wave  each  be  resolved  into  a  sine  and  a 
cosine  component  referred  to  a  as  the  origin  from  which  to  reckon 
time.  Then  the  equation  of  the  wave  is 


y  =  Ci  sin  (ut  + 
=  A  i  sin  w<  + 

-f 


0  +  C3  sin 

1  cos  cot 

s  sin  3w£  + 


03) 


cos 


(33) 

(34) 


Where 


+ 


tan  0i  = 


tan  33  = 


Divide  the  whole  wave  length  between  a  and  b  into  three  equal 
parts  by  three  equally  spaced  ordinates  shown  in  Fig.  28.  Call 
the  lengths  of  these  ordinates  YSi,  F32  and  YBS. 

At  the  origin  t  =  0  and  each  sine  component  of  the  wave 
(see  equation  (34))  is  zero,  but  each  cosine  component  has  its 
maximum  value.  The  length  of  the  ordinate  erected  at  the  point 
t  =  0  will  therefore  be  equal  to  the  sum  of  the  cosine  terms  of  the 
fundamental  and  harmonics. 


F3i  =  B,  +  B, 


96  PRINCIPLES  OF  ALTERNATING  CURRENTS  . 

In  general,  if  there  are  k  harmonics,  the  ordinate  erected  at 
the  point  t  =  0  will  be  equal  to  the  sum  of  the  maximum  values 
of  the  cosine  terms  for  the  fundamental  and  all  harmonics. 

Yml  =  Bl  +  £3  +  £5  +  .  .  .  +  Bk  (35) 

Referring  again  to  equation  (34)  and  Fig.  28,  it  is  obvious 
from  equations  (29)  and  (32)  that  if  F3i,  F32  and  F33  are  the 
lengths  of  the  ordinates  at  the  points  1,  2  and  3, 

i(F31  +  F32  +  F33)  =  B3 

or  the  maximum  value  of  the  cosine  component  for  the  third 
harmonic. 

If  the  wave  contained  harmonics  which  were  multiples  of  the 
third,  one  third  of  the  sum  of  the  ordinates  F3i,  F32  and  F33 
would  be  the  sum  of  the  maximum  values  of  the  cosine  terms  for 
the  third  and  all  harmonics  which  were  multiples  of  the  third. 
That  is,  in  general 

^(F31  +  732  +  F33)  =  (#3  +  Bg  +  £15  +  etc.)  (36) 

o 

Similarly  if  a  wave  containing  harmonics  is  divided  into  five 
equal  parts  by  five  equally  spaced  ordinates,  F51,  F52,  F53, 
F54  and  F55,  the  first  ordinate  being  erected  at  the  point  a, 

*(F5i  +  F52  +  F53  +  F54  +  F55)  =  (B6  +  Bn  +  #25  +  etc.) 
o 

Dividing  the  wave  into  seven  and  into  nine  parts  gives 

(F71  +  772  +   •    •    •   +  F77)  =  (B7  +  B2l  +  £35  +  etc.) 


92 


F99)  =  (B9  +  £27  +  £45  +  etc.) 


In  practice  it  is  convenient  to  erect  the  first  ordinate  at  the 
point  where  the  curve  to  be  analysed  crosses  the  axis  of  time. 
In  this  case  Ym\  is  zero  and  from  equation  (35) 

Bl  +  Bz  +  B5  +  etc.  =  0 
Bi  =  -  £3  -  B5  -  etc. 

For  most  current  and  voltage  waves  met  in  practice  harmonics 
of  higher  order  than  the  seventh  are  not  important.  Relatively 
high  harmonics  may  sometimes  be  produced  by  the  armature 


NON-SINUSOIDAL  WAVES  97 

teeth  of  motors  or  generators.  If  present  in  any  appreciable 
magnitude  they  are  as  a  rule  easily  detected  by  inspection  of  the 
wave  and  may  then  be  calculated. 

If  harmonics  of  higher  order  than  the  seventh  are  negligible 

#1  =  --  £3  -  B5  -  B7  (37) 

£3  =  i(F31  +  F32  +  F33)  (38) 


F52  +    .    .    .    +  F55)  (39) 

B7  =  i(F71  +  F72  +    .    .    .    +  F77)  (40) 

If  the  ninth  harmonic  were  present  as  well  as  the  third,  fifth 
and  seventh 


F32  +  F33)  (41) 


£9  =    (F91  +  F92  +  .    .    .    .    .    .+  F99)  (42) 

Bs  =  |(F31  +  F  32  +  F33)  -  B9  (43) 

When  the  approximate  equations  (equations  (37)  to  (40) 
inclusive)  are  used  to  analyse  a  wave,  the  base  line  may  be  ap- 
propriately divided  to  detect  higher  harmonics.  If  they  exist 
in  appreciable  magnitude  they  must  be  corrected  for  by  the 
method  indicated  for  correcting  the  third  for  the  presence  of  the 
ninth.  (See  equation  (43).) 

Each  of  the  sine  terms  in  the  wave  has  its  maximum  value 
one  quarter  of  a  period  (measured  on  the  scale  of  angles  for  the 
harmonic  considered)  from  the  initial  ordinate,  Ym\,  or  in  Fig. 
28,  page  95,  from  the  ordinate  F31.  If  the  ordinates  on  Fig.  28 
are  shifted  in  the  direction  of  lag  (to  the  right)  one  quarter  of  a 
period  for  the  third  harmonic,  one  third  of  their  sum  will  be  equal 
to  the  maximum  value  of  the  sine  term  for  the  third  harmonic. 

|(F31'  +  F32'  +  F33')  =  A3  (44) 

The  primes  on  the  F's  indicate  that  they  have  been  shifted 
in  the  direction  of  lag  from  their  original  position,  i.e.,  from 
the  position  in  which  they  were  drawn  for  determining  the  coeffi- 


98  PRINCIPLES  OF  ALTERNATING  CURRENTS 

cients  of  the  cosine  terms,  by  one-quarter  of  a  period  for  the 
harmonic  considered. 

If  the  wave  contains  harmonics  which  are  multiples  of  the 
third, 

£(1V  +  F32'  +  F33')  =  (A,  -  A9  H-  A15  -  etc.)       (45) 

o 

Similarly,  if  the  F5  group  of  ordinates  for  the  fifth  harmonic 
is  shifted  in  the  direction  of  lag  by  one-quarter  of  a  period  for  the 

fifth  harmonic,  i.e.,  by  -=-  degrees  or  -=  X  ~  radians  on  the  funda- 
o  o        z 

mental  scale  of  angles, 

^(F51'+F52'  +  .   •   •   +  F55')  =  (A5-A15-fA25-etc.)   (46) 

If  the  Y7  group  of  ordinates  are  shifted  in  the  direction  of  lag 
by  one-quarter  of  a  period  for  the  seventh  harmonic, 

i(F71'  +  IV  +  .   .   .  +  F77')  =  (A7-A21  +  A35-etc.)  (47) 
If  the  harmonics  above  the  seventh  are  negligible, 

A3  -  *(F31'  +  F32'  +  F33')  (48) 

A5  =  ^OV  +  F52'+.    .    .+  F55')  (49) 

A1=l(Y71'  +  Yn'  +  .    .    .+  F77')  (50) 

If  the  ninth  harmonic  is  also  present,  A3  as  found  by  equation 
(48)  must  be  corrected  for  that  harmonic.  For  example: 

A9  =-g(F9i'+  F92'+.    .   .  +  IV)  (51) 

A3  =  i(r81;  +  F32r  +  F330  +  A9  (52) 

If  other  high  order  harmonics  are  present,  the  other  coefficients, 
A  5  and  A7,  must  similarly  be  corrected.  The  method  of  correc- 
tion is  the  same  as  indicated  for  the  coefficients  A3  and  Bs. 

If  the  first  ordinate,  Ym\,  of  the  group  of  ordinates  for  the 
cosine  series  is  shifted  in  the  direction  of  lag,  one-quarter  of  a 
period  for  the  fundamental,  its  length  will  be 

Yml'  =  A!  -  A3  +  A5  -  A7  +  A9  -  etc. 
A!  =  Yml'  -  (-  A3  +  A5  -  A7  +  A9  -  etc.)        (53) 


NON-SINUSOIDAJ.  WAVES 
The  even  harmonics  do  not  contribute  to  the  ordinate  Y, 


99 


That  equation  (53)  is  true  when  either  odd  or  even  harmonics 
or  both  are  present  will  be  understood  when  the  phase  relations 
among  the  fundamental  and  harmonics  of  the  sine  -terms  are 
considered  at  the  ordinate  Ymi'. 

A  change  of  a  degrees  in  the  position  of  any  ordinate  measured 
on  the  scale  of  angles  for  the  fundamental,  is  equivalent  to  a 
change  in  position  of  ka  degrees  with  respect  to  the  kih  harmonic 
scale  of  angles.  The  following  table  is  based  on  this  relationship. 
This  table  gives  the  change  in  phase  produced  in  each  harmonic 
by  moving  the  axis  of  reference  in  the  direction  of  lag  through 
one-quarter  of  a  period  for  the  fundamental. 


•  i  tl  I  11J.VS11.II> 

^       lUlllJi/        111     LJilCfcO^                                1   1  i.l  1    1  I  1  '  M  1  l< 

V      MilJl^v        All      |MUl.^' 

Fundamental 

90° 

8th 

8  X  90°  =    720° 

o        0° 

2nd                 2  X  90°  =  180° 

9th 

9  X  90°  =     810C 

o      90° 

3rd 

3  X  90°  =  270° 

10th 

10  X  90°  =    900C 

=c=    180° 

4th 

4  X  90°  =  360° 

llth 

11  X  90°  =    990° 

o      0° 

o    270° 

5th 

5  X  90°  =  450° 

12th 

12  X  90°  =  1080° 

o    90° 

o        0° 

6th 

6  X  90°  =  540° 

13th 

13  X  90°  =  1170C 

o  180° 

o      90° 

7th 

7  X  90°  =  630° 

14th 

14  X  90°  =  1260° 

o  270° 

o    180° 

The  cosine  terms  for  the  fundamental  and  all  harmonics  are 
in  phase  at  the  ordinate  Ymi  and  add  directly  to  it.  The  sine 
terms  for  the  fundamental  and  all  harmonics  are  zero  at  the 
ordinate  Ymi  and  contribute  nothing  to  its  magnitude.  Accord- 
ing to  the  table,  the  fifth,  ninth,  thirteenth,  etc.  harmonics  of 
the  sine  series  will  have  their  maximum  values  at  a  point  90 
degrees  from  the  ordinate  Fml,  measured  in  the  direction  of  lag 
on  the  fundamental  scale  of  angles,  and  will  add  directly  to 
the  maximum  value  of  the  sine  term  of  the  fundamental  at  an 
ordinate,  Fm/,  erected  90  fundamental  degrees  in  the  direction  of 
lag  from  the  ordinate  Yml.  A  shift  in  the  position  of  the  axis  of 


100 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


90  fundamental  degrees  in  the  direction  of  lag  changes  the  phase 
of  the  third,  seventh,  eleventh,  etc.  harmonics  in  the  direction 
of  lag  by  270  degrees,  measured  on  their  own  scales  of  angles,  or 
by  360  —  -270  =  90  degrees  in  the  direction  of  lead.  They  will 
therefore  be  opposite  in  phase  to  the  sine  terms  of  the  fundamen- 
tal, the  fifth,  ninth,  thirteenth,  etc.  harmonics  at  the  ordinate 
Ym\  and  will  subtract  from  it. 

The  even  harmonics  of  the  sine  terms  are  all  changed  in  phase 
by  either  180  or  0  degrees  by  a  shift  in  the  reference  axis  of  90 
fundamental  degrees  in  the  direction  of  lag.  They  will  conse- 
quently all  have  zero  values  at  the  ordinate  Ymi,  as  well  as  at 
the  ordinate  Ymi,  and  will  contribute  nothing  to  it. 

The  cosine  terms  for  the  fundamental  and  all  harmonics  will 
have  zero  values  at  the  ordinate  Ymi  and  will  contribute  nothing 
to  it. 


FIG.  29. 

Most  current  and  voltage  waves  met  in  practice  are  symmetri- 
cal, i.e.,  their  positive  and  negative  loops  are  identical  except  in 
sign.  When  the  positive  and  negative  loops  of  a  wave  are 
symmetrical,  it  is  necessary  to  construct  only  the  first  loop,  as 
the  length  of  any  given  ordinate  for  the  second  loop  may  be 
determined  from  a  suitably  placed  ordinate  in  the  first  loop. 

Consider  a  third  harmonic.     Refer  to  Fig.  29. 

Let  the  three  ordinates  marked  1/2  and  3  at  their  upper  ends 
be  the  ordinates  for  determining  one  component,  such  as  the 
cosine  component-,  of  the  third  harmonic.  If  there  are  no 
harmonics  present  which  are  multiples  of  the  third 

B,  =  V»i  +  Y32  +  F33) 


NON-SINUSOIDAL  WAVES  MGl. 

Instead  of  dividing  the  whole  wave  into  k  parts  (in  this  case 
three),  where  k  is  the  order  of  the  harmonic  to  be  determined, 
let  the  half  wave  be  divided  into  k  parts,  as  shown  in  Fig.  29. 
The  numbers  for  the  k  ordinates  for  the  half  wave  or  the  2k 
ordinates  for  the  whole  wave  have  circles  around  them  and  are 
placed  at  the  bottom  of  the  ordinates  in  Fig.  29. 

The  second  of  the  six  ordinates,  i.e.,  of  the  2k  ordinates, 
is  obviously  equal  in  magnitude  but  opposite  in  sign  to  the  third 
of  the  three  ordinates  marked  at  their  upper  ends.  Therefore 

£3  =  |(F.i  -  F62  +  F63)  (54) 

If  in  addition  to  the  third  harmonic  the  wave  contains  har- 
monics which  are  multiples  of  the  third 

B3  +  B9  +  £15  +  etc.  =  *  (F61  -  762  +  F63)         (55) 

To  determine  the  fifth  harmonic,  divide  the  half  wave  into 
five  parts.  This  corresponds  to  dividing  the  whole  wave  into 
ten  parts.  Then 

Bb  +  £15  +  #25  +  etc.  =  i(Fio  i  -  F10  2  +  F10  3  -  FIO 4  +  F 10  5)  (56) 

Similarly,  by  dividing  the  half  wave  into  seven  parts,  the 
seventh  harmonic  plus  the  others  of  its  group  may  be  found. 
Dividing  the  half  wave  into  seven  parts  corresponds  to  dividing 
the  whole  wave  into  fourteen  parts. 

B7  +  £21  +  B35  +  etc.  =  i(714  l  -  Yu  2  +  Ylt  , 

-  Fi4  4  +  F14  5  -  Fi4  e  +  Fu  r)      (57) 

When  the  loop  of  a  symmetrical  wave  is  divided  into  k  parts 
by  equally  spaced  ordinates,  to  determine  the  kth  harmonic  the 

k  —  1 
sum  of  the  — ^ —  ordinates  with  even  numbers  is  subtracted  from 

k  +  1 
the  sum  of  the  — ^ —  ordinates  with  odd  numbers. 

Zi 

It  is  convenient,  when  analyzing  a  wave  containing  only  odd 
harmonics,  to  divide  the  half  wave  into  2k  equal  parts  by  2k 
equally  spaced  ordinates,  erecting  the  first  ordinate  where  the 
wave  crosses  the  axis  of  time.  When  2k  ordinates  are  thus  drawn 
in  the  half  wave,  those  with  odd  numbers  are  used  for  determining 


102  0 


OF  ALTERNATING  CURRENTS 


the  coefficients  of  the  cosine  terms.  Those  with  even  numbers 
will  be  one  quarter  period  (for  the  fcth  harmonic)  from  the  others, 
and  are  used  for  determining  the  coefficients  of  the  sine  terms. 
For  example,  suppose  a  wave  contains  a  fifth  harmonic  but  no 
others,  such  as  the  fifteenth,  which  would  be  included  with  the 
fifth.  Let  the  half  wave  be  divided  into  ten  parts  and  number  the 
ordinates  Yl}  Y2,  Fs,  .  .  .  F10.  Then 


\  -  Y,  +  F5  -  F7  +  F9) 


As  =  d 


-  F4  +  F6  - 


F10) 


(58) 
(59) 


That  is,  the  ordinates  with  odd  numbers  are  used  with  alternate 
signs  to  determine  the  B  coefficients,  and  the  ordinates  with  even 
numbers  are  used  with  alternate  signs  to  determine  the  A  co- 
efficients. In  each  case  the  first  ordinate  is  considered  positive. 

An  analysis  of  a  wave  containing  only  odd  harmonics  will  make 
this  clear. 

When  calculating  the  B  and  A  coefficients  of  a  wave  by  the 
Fischer-Hinnen  method,  it  must  be  remembered  that  the  coeffi- 
cients are  not  obtained  separately  but  in  groups.  For  this  reason, 
it  is  always  necessary  to  determine,  by  an  approximate  division  of 
the  wave,  the  highest  order  harmonic  that  is  present  in  appreciable 
magnitude,  in  order  that  the  coefficients  given  by  the  approximate 
equations  (37)  to  (40)  and  (48)  to  (50)  inclusive  may  be  corrected 
for  the  presence  of  high  order  harmonics. 


FIG.  30. 


Example  of  the  Analysis  of  a  Wave  Containing  Only  Odd 
Harmonics  by  the  Fischer-Hinnen  Method.— The  wave  of  the 
magnetizing  current  of  a  60-cycle  transformer  will  be  analyzed. 
Analysis  shows  that  this  wave  contains  a  pronounced  third 


NON-SINUSOIDAL  WAVES 


103 


harmonic,  a  fairly  large  fifth  harmonic  and  small  seventh  and 
ninth  harmonics.     The  harmonics  above  the  ninth  are  negligible. 

The  half  wave,  with  the  proper  ordinates  for  determining  the 
coefficients  for  the  third  harmonic,  is  shown  in  Fig.  30.  Since 
the  ninth  harmonic  is  not  negligible,  these  coefficients  must  be 
corrected  for  its  presence. 

Measurements  made  on  the  original  curve  from  which  Fig.  30 
was  reproduced  gave  the  following  lengths  of  the  ordinates  in 
inches: 

F!  =  0.00         F2  =  0.99  F3  =  1.80  F4  =  1.01 

F5  =  0.51         F6  =  0.30 

0.51  1.29  1.80  1.01 

The  scale  for  the  ordinates  of  the  original  curve  was  one  ampere 
per  inch. 

£3  +  £9  =    KFi  -  F3  +  F5) 
=    K0.51  -  1.80) 
=  —0.43  inches 

=  -0.43  X  1  =  -  0.43  ampere. 
A,  +  A9  =    KF2  -  F4  +  F6) 
=    ±(1-29  -  1.01) 
=      0.093  inches 
=      0.093  X  1  =  0.093  ampere. 

The  half  wave  with  the  proper  ordinates  for  finding  the 
fifth  harmonic  is  shown  in  Fig.  31. 

23456789  10 


FIG.  31. 

Measurements  made  on  the  curve  from  which  Fig.  31   was 
reproduced  gave  the  following  lengths  of  the* ordinates  in  inches. 
Yl  =  0.00          Y2  =  0.51         F3  =  1.32         F4  =  1.82 
F5  =  1.52          F6  =  1.01         FT  =  0.65         F8  =  0.44 
F9  =  0.33        F10  =  0.22 


1.85 


1.74 


1.97 


2.26 


104  PRINCIPLES  OF  ALTERNATING  CURRENTS 


i  -  Y,  +  F5  -  Y7  +  Y9) 
=    i(1.85  -  1.97) 
=  -0.024  inches 
=  -0.024  X  1  =  -0.024  ampere. 

A5  =  i(F2  -  F4  +  F6  -  F8  +  710) 

=  i(1.74  -  2.26) 

=  -0.10  inches 

=  -0.10  X  1  =  -0.10  ampere. 

Further  analysis  shows  that 

B7  =  0.027  ampere. 
A7  =  0.003  ampere. 
BQ  =  0.004  ampere. 
A  9  =  0.014  ampere. 

£3    =    (#3   +   £9)    ~   #9 

=  -0.43  -  0.004  =  -0.43  ampere. 
A,  =  (A8-A9)  +  A, 

=  0.093  +  0.014  =  0.107  ampere. 

From  equation  (37),  page  97 
Bl  =  -(B3  +  £5  +  #7  +  #9) 

=  -{(-0.43)  +  (-0.024)  +  (0.027)  +  (0.004)] 
=  0.42  ampere. 

From  equation  (53),  page  98 

Al  =  Fml'  -  (-^3  +  A,  -  A7  +  A9) 

By  measurement  on  the  original  figure,  the  ordinate  4 
(see  Fig.  30),  which  is  displaced  ninety  fundamental  degrees 
in  the  direction  of  lag  from  the  ordinate  marked  1,  is  equal  to 
1.01  amperes. 

Al  =  1.01  -  {  -(0.107)  +  (-0.10)  -  (0.003)  +  (0.014)) 
=  1.21  amperes. 

i  —  A  i  sin  ut  +  BI  cos  ut 

+  A  3  sin  3ut  +  Bz  cos  3o>£ 
+  A  5  sin  5co£  -f-  B5  cos  5co£ 
+  A7  sin  7co£  +  £7  cos  7co£ 
+  A  9  sin  9o>Z  +  £9  cos  9co£ 


NON-SINUSOIDAL  WAVES  105 

1.21  sin  377*  +  0.42  cos  377* 
+  0.107  sin  1131*  -  0.43  cos  1131* 
-  0.10  sin  1885*  -  0.024  cos  1885* 
-f  0.003  sin  2639*  +  0.027  cos  2639*' 
+  0.014  sin  3393*  +  0.004  cos  3393* 


i)2  +  (o.42)2 

=  1.28  amperes. 

"'=li  =  rf  =  °-347 

0i  =  +19.1  degrees. 


(  -0.43)2 
=  0.443  ampere. 


I93  =  -  76.0  degrees. 

C6  =  VA52  +  ^52  =  V(-0.10)2  +  (-0.024) 
=  0.103  ampere. 


05  =  —166.5  degrees. 

C7  =  V^72  +  B72  =  V(0.003)2  +  (0.027) 
=  0.027  ampere. 


. 

AT       0.003 

=  +84  degrees. 

£92  =  V  (0.014)  2+(0.004)" 
=  0.0146  ampere. 


09  =  +16  degrees. 

1.28  sin  (377*  +  19?1)  +  0.443  sin  (1131*  -  76?0) 
+  0.103  sin  (1885*  -  166°5)  +  0.027  sin  (2639*  +  84°) 
+  0.015  sin  (3393*  +  16°) 


106 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Form  Factor. — The  form  factor  for  an  alternating  current  or 
voltage  is  the  ratio  of  its  effective  or  root-mean-square  value 
to  its  average  value. 


Form  factor  = 


Effective  value 
Average  value 


(60) 


For  a  sinusoidal  wave  the  form  factor  is 

Em^ 

r» 

=  1.11 


2E, 


The  form  factor  is  usually  less  than  1.11  for  a  flat-topped 
wave,  i.e.,  a  wave  that  is  flatter  than  a  sinusoidal  wave.  It  is 
usually  greater  than  1.11  for  a  peaked  wave,  i.e.,  one  that  is  more 

peaked  than  a  sinusoidal 
wave. 

The  form  factor  has 
not  much  practical  im- 
portance, as  two  waves 
having  totally  different 
wave  forms  may  have 
equal  form  factors.  For 
example,  the  following 
wave  has  the  same  form 
factor  as  a  sinusoidal  wave  but  it  is  of  totally  different  wave 
shape. 


FIG.  32. 


mi  sin  u>t  +  -  Imi  sin  3wt 

This  wave  and  a  sine  wave  having  the  same  root-mean-square 
or  effective  value  are  plotted  in  Fig.  32. 

Amplitude,  Crest  or  Peak  Factor.  —  The  amplitude,  crest 
or  peak  factor  of  an  alternating  wave  is  the  ratio  of  its  maximum 
value  to  its  root-mean-square  or  effective  value.  A  knowledge 
of  the  amplitude  or  peak  factor  is  of  importance  in  testing  insula- 
tion, since  the  stress  to  which  an  insulation  is  subjected  by  a 
given  impressed  voltage  depends  upon  the  maximum  value  of  the 


NON-SINUSOIDAL  WAVES  107 

voltage  and  not  upon  its  root-mean-square  or  effective  value. 
A  knowledge  of  the  form  factor  of  a  voltage  in  connection  with 
insulation  testing,  however,  is  of  little  if  any  value.  The  peak 
factor  of  a  sinusoidal  wave  is  -\/2  =  1.41. 

Deviation  Factor. — The  deviation  factor  of  a  wave  is  the 
ratio  of  the  maximum  difference  between  the  corresponding 
ordinates  of  the  actual  wave  and  an  equivalent  sine  wave  of  equal 
length,  to  the  maximum  ordinate  of  the  equivalent  sine  wave, 
when  the  waves  are  superposed  in  such  a  way  as  to  make  the 
maximum  difference  between  corresponding  ordinates  as  small  as 
possible.  (For  definition  of  equivalent  sine  wave  see  the  next 
paragraph.)  Except  in  special  cases,  a  deviation  factor  of  0.10  is 
permissible  in  the  wave  form  of  commercial  electrical  machinery. 

Equivalent  Sine  Waves. — In  most  of  the  alternating-current 
phenomena  which  are  met  in  practice,  neither  the  voltage  nor  the 
current  is  sinusoidal,  although  both  are  periodic.  In  many  cases 
where  the  wave  forms  do  not  differ  greatly  from  a  sinusoid,  it  is 
sufficiently  exact  to  replace  the  non-sinusoidal  waves  of  voltage 
and  current  by  equivalent  sine  waves.  These  equivalent  sine 
waves  have  the  same  root-mean-square  or  effective  values  as  the 
actual  waves  they  replace  and  their  phase  is  so  chosen  as  to 
make  El  cos  6  for  the  equivalent  sine  waves  equal  to  the  actual 
power.  The  sign  of  the  phase  angle  between  the  equivalent 
sine  waves  is  made  the  same  as  that  of  the  phase  angle  between 
the  fundamentals  of  the  actual  waves.  When  the  fundamentals 
are  in  phase,  but  the  power-factor  is  not  unity,  the  sign  of  the 
equivalent  phase  angle  is  indeterminate. 

Complex  waves,  which  differ  much  in  wave  form,  cannot  be 
replaced  by  their  equivalent  sine  waves,  when  they  are  to  be  added 
or  subtracted,  without  danger  of  introducing  considerable  error. 
(See  page  110.) 

Equivalent  Phase  Difference. — The  angle  0,  determined  by 

p 
0  =  cos"1  j^j  when  voltage  and  current  are  not  sinusoidal  waves, 

is  known  as  the  equivalent  phase  angle.  I  and  E  are  the  effec- 
tive or  root-mean-square  values  of  the  non-sinusoidal  waves  andP 
is  the  average  power.  The  equivalent  phase  difference  is  the  phase 
angle  that  must  be  used  with  the  equivalent  sine  waves  of  current 
and  voltage  to  produce  the  true  power.  The  equivalent  phase 


108  PRINCIPLES  OF  ALTERNATING  CURRENTS 

difference  may  often  be  misleading,  since  the  presence  of  har- 
monics in  one  wave  which  are  not  present  in  the  other  will  lower 
the  power-factor  even  though  there  is  no  displacement  between 

p 
the  waves.     In  general  6,  as  determined  by  0  =  cos"1  -^  has  no 

real  physical  significance  except  when  sine  waves  are  considered. 
Example  of  Equivalent  Sine  Waves. — The  analysis  of  the 

voltage  and  current  waves  for  a  certain  circuit  has  shown  them  to 

be  of  the  forms 

e  =  100  sin  ut  +  20  sin  (3co£  +  60°)  + 15  sin  (5^-40°) 
i  =  40  sin  («*  -  30°)  +  5  sin  (3co£  +  20°) 

E  .  ^(ioo)V+  («))'  ±W  ,  72>9  volts 

1(40) 2  +  (5)2 
/  =  A/—  -  =  28.5  amperes 


P  =        - cos  30°  +  -  cos  40" 

2t  2 

=  1770  watts 
Power-factor  =  m^5  =  0.852 

Equivalent  phase  difference  =  6  =  cos"1  0.852 

=  31.6  degrees. 

If  the  equivalent  sine  voltage  is  taken  zero  when  time,  t,  is 
zero,  the  equivalent  sine  waves  of  voltage  and  current  are 

e  =  v"2  x  72.9  sin  ut 

=  103.1  sin  cot 
i  =  \/2  X  28.5  sin  (co*  -  31°6) 

=  40.3  sin  (co*  -  31°6) 

If  the  equivalent  sine  current  is  taken  zero  when  time,  t, 
is  zero,  the  equivalent  sine  waves  are 

e  =  103.1  sin  (ut  +  31?6) 
i  =  40.3  sin  ut 

It  should  be  noticed  that  the  fifth  harmonic  in  the  voltage 
contributes  nothing  to  the  power,  since  there  is  no  component  in 
the  current  of  the  same  frequency.  Although  it  contributes 
nothing  to  power  it  does  increase  the  voltage  and  therefore 


NON-SINUSOIDAL  WAVES  109 

lowers  the  power-factor.  Because  the  fifth  harmonic  contributes 
nothing  to  power,  it  cannot  be  neglected  in  finding  the  root- 
mean-square  or  effective  value  of  the  voltage. 

As  has  already  been  stated,  the  power-factor  of  a  circuit 
cannot  be  unity  unless  the  current  and  voltage  contain  like 
harmonics,  and  then  the  relative  magnitudes  and  the  phase  rela- 
tions of  the  harmonics  must  be  identical  in  the  two  waves.  The 
power-factor  of  a  circuit  containing  nothing  but  pure  resistance 
cannot  be  unity,  even  though  in  such  a  circuit  there  would  be 
no  displacement  between  the  current  and  voltage  waves,  unless 
the  temperature  coefficient  of  the  material  of  which  the  resistance 
is  made  is  zero  or  the  dimensions  of  the  resistance  unit  are  such 
that  there  is  no  appreciable  change  in  its  resistance  during 
a  cycle.  If  a  resistance  unit  is  made  of  fine  wire  of  high  tempera- 
ture coefficient  its  resistance  will  change  appreciably  during  a 
cycle.  If  a  sinusoidal  voltage  is  impressed  on  such  a  resistance 
the  current  will  be  flatter  than  a  sinusoid  and  will  therefore  con- 
tain harmonics,  among  which  will  be  a  marked  third.  The 
power-factor  of  a  circuit  whose  resistance  varies  with  current 
during  a  cycle  cannot  be  unity,  even  though  the  circuit  con- 
tains nothing  but  pure  resistance.  For  commercial  circuits,  the 
change  of  the  resistance  with  current  during  a  cycle  is  usually 
too  small  to  produce  any  noticeable  effect  on  the  shape  of  the 
current  wave. 

Consider  a  case  where  there  is  no  phase  displacement  between 
the  current  and  voltage  waves  but  the  current  wave  contains  a 
harmonic  which  is  not  present  in  the  voltage.     Let 
6  =  100  sin  2x60* 
i  =  10  sin  2ir6Qt  +  5  sin  6rr60< 

The  voltage  is  a  pure  sinusoidal  wave,  while  the  current  contains 
a  50  per  cent,  third  harmonic.  This  third  harmonic  contributes 
nothing  to  the  power,  since  the  voltage  has  no  third  harmonic. 
The  root-mean-square  or  effective  value  of  the  voltage  and  the 
current  are  respectively 


=  70.7  volts 


110  PRINCIPLES  OF  ALTERNATING  CURRENTS 

P  =  1Q°  *  10  cos  (0°  -  0°)  +  0 

2 

=  500  watts 
Power-factor  >  ^^  .  0.895 

Equivalent  phase  difference  =  0  =  cos~1  0.895=  ±26.5  degrees. 
The  equivalent  sine  waves  are 

e  =  100  sin  2^60* 

i  =  T/2  X  7.91  sin  (27r60*  ±  26?5) 
=  11.18  sin  (27r60*  ±  26?5) 

Although  the  power  in  the  example  just  given  is  due  entirely 
to  the  sinusoidal  voltage,  e  =  100  sin  2^60*,  and  the  sinusoidal 
current,  i  =  10  sin  2?r60*,  the  latter  is  not  the  equivalent  sine 
wave  of  current  since  it  does  not  have  the  proper  ampere  value 
or  phase  relation. 

Where  an  exact  analysis  of  any  particular  problem  is  essential, 
the  substitution  of  the  equivalent  sine  waves  for  the  actual  volt- 
age and  current  is  not  permissible. 

Addition  and  Subtraction  of  Non-sinusoidal  Waves. — When 
non-sinusoidal  currents  or  voltages  are  to  be  added  or  subtracted, 
each  must  first  be  expressed  in  terms  of  its  Fourier  series,  i.e.,  in 
terms  of  its  fundamental  and  harmonics.  The  fundamentals  and 
the  harmonics  of  like  frequency  may  then  be  added  or  subtracted 
vectorially  to  give  the  fundamental  and  the  harmonics  of  the  re- 
sultant wave.  Equivalent  sine  waves  cannot  be  added  or  sub- 
tracted vectorially  except  when  the  wave  forms  are  identical  and 
the  phase  displacement  is  zero.  If  the  wave  forms  are  very 
different  or  the  phase  displacement  between  the  waves  is  great, 
the  error  produced  by  adding  or  subtracting  the  equivalent  sine 
waves  may  be  large. 

Example  of  Addition  of  Non-sinusoidal  Waves. — Let  the 
following  voltage  be  impressed  on  a  circuit  having  two  branches 
in  parallel,  one  consisting  of  a  resistance  of  5  ohms  in  series  with  a 
condenser  of  132.7  microfarads  capacitance,  the  other  consisting 
of  10  ohms  resistance  in  series  with  an  inductance  of  0.0398 
henry. 

e  =  200  sin  (377*  +  10°)  +  75  sin  (1131*  +  30°) 

+  50  sin  (1885*  +  50°) 


NON-SINUSOIDAL  WAVES  111 

The  currents  in  the  two  branches  will  be  (See  Chapter  VII) 

i'  =  9.71  sin  (377*  +  85°96) 

+  9.00  sin  (1131*  +  83?15) 

+  7.81  sin  (1885*  +  88?66) 
i"  =11.09  sin  (377*  -  46?31) 

+  1.625  sin  (1131*  -  47?47) 

4-  0.661  sin  (1885*  -  32°40) 

Capital  letters  with  the  subscripts  1,  3  and  5  will  represent 
root-mean-square  values  of  the  fundamental  and  harmonics. 
The  subscript  0  used  with  the  subscripts  1,  3  and  5  will  indicate 
resultants. 

Consider  the  vectors  representing  the  fundamentals  of  the 
currents  at  the  instant  when  time,  *,  is  zero.  Their  vector 
expressions  are 

\/2  //  =  9.71(cos  85°96  +  j  sin  85?96) 

=  0.684  +  J9.68 
\/2/i"=  11.09(cos46°31  -  j  sin  46?31) 

=  7.67  -  ./8.03 
\/2  loi  =  8.35  -f  jl.65 
V2  7oi  =  V (8.35)  2  +  (1.65)2  =  8.51  amperes  maximum. 

tan  0oi  =  ^^  =  0.1976        00i  =  11.18  degrees. 

o.oO 


Consider  the  vectors  representing  the  third  harmonics  at  the 
instant  *  is  zero. 


\/2  73'  =  9.00(cos  83°15  +j  sin  83°15) 

=  1.074  +J8.94 
V2  h"  =  1.625(cos  47?47  -  j  sin  47?47) 

=  1.098  -  j  1.198 
V2  7as  =  2.172+/7.74 

^  =  \/(2.172)2  +  (7.74)  2  =  8.04  amperes  maximum. 


tan  003  =      '        =  3.563       003  =  74.32  degrees. 


112  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Consider  the  vectors  representing  the  fifth  harmonics  at  the 
instant  t  is  zero. 

\/2  I*'  =  7.81  (cos  88°66  +  j  sin  88°66) 

=  0.183  +/7.81 
\/2  75"  =  0.661  (cos  32°40  -  j  sin  32?40) 

=  0.558  -  jO.354 
\/2  705  =  0.741  +  J7.46 
\/2  705  =  V(0.741)2  +  (7.46) 2  =  7.50  amperes  maximum. 

7  46 
tan  005  =  0~7^j  =  10.06       005  =  84.32  degrees. 

The  resultant  current  wave  is 

to   =  8.51  sin  (377*  +  11?18)  +  8.04  sin  (1131*  +  74°32) 

+7.50  sin  (1885*  +  84?32) 

7  /(8.51)2+  (8.04) 2  +"(7T5Q)'2 

/o  =  \/-  — «^  =  9.84  amperes  effective. 

The  resultant  power  is 

Po  =  20°  ^  8'51  cos  (10°  -  11?18) 

+  75  X  f '°4  cos  (30°  -  74?32) 


5°  X7'5°  cos  (50°  -  84?32) 


=  851  +  216  +  155  =  1222  watts. 
The  power  in  each  branch  of  the  parallel  circuit  is 
P'=  200X9,71  C0g(_75o96) 


235.4  +  202.3  +  152.4  =  590.1  watts. 
200  X  11.09  cos(+56?31) 


=  615.2  +  13.2  +  2.2  =  630.6  watts 
P0  =P'+P"  =  590  +  631 
=  1221  watts 


NON-SINUSOIDAL  WAVES  113 

The  maximum  value  of  the  equivalent  sine  wave  of  voltage  is 
=  V(200)2  +  (75)2  +  (50)2  =  219.3  volts  maximum. 


The  maximum  values  of  the  equivalent  sine  waves  of  the  cur- 
rents are 


=  \/(9.71)2  +  (9.00)2  +  (7.81)2     :  15.37     amperes 
maximum. 


V27"  =  Vai.09)2  +  (1.625)2  +  (0.661)2  =  11.23  amperes 
maximum. 


(Power-factor)'  =  =  0.3501 


V2  '      V2 

Equivalent  phase  difference  =  0'  =  cos"1  0.3501  =69.50  degrees. 
(Power-f actor) "  =  oino630'^  00  =  0.5121 

Equivalent  phase  difference  =  9"  =  cos"1 0.5121  =59.20  degrees. 
If  the  equivalent  sine  voltage  is  taken  zero  when  t  is  zero,  the 
equivalent  sine  waves  are 

e    =  219.3  sin  377* 

if    =  15.37  sin  (377*  +  69?50) 

i"  =  11. 23  sin  (377*  -  59?20) 

Add  the  equivalent  sine  currents  as  if  they  were  actually 
sinusoidal  waves.  Consider  the  vectors  representing  them  at  the 
instant  *  is  zero. 

\/27'     =  15.37  (cos  69?50  +  j  sin  69?50) 

=  5.38  +  J14.4 
V27"    =  11.23  (cos  59°20  -  j  sin  59?20) 

=  5.73  -  J9.64 
V2/o     =  11.11  +  J4.76 
\/27o     =  V(H.ll)2  +  (4.76)2  =  12.09  amperes  maximum. 

12  09 
70   =  — 'j=-  =  8.56  amperes  effective. 

8 


114  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Adding  the  equivalent  sine  waves  vectorially,  in  the  example 
just  given,  gives  8.56  amperes  for  the  resultant  current  instead 
of  9.84  amperes,  the  correct  value.  The  error  is  13  per  cent. 
If  both  branches  of  the  divided  circuit  had  contained  similar 
constants,  the  error  of  adding  the  equivalent  sine  waves  would 
probably  have  been  less.  It  could  not  have  been  zero  unless 
the  wave  forms  were  identical  and  the  two  waves  were  in  phase. 

For  example,  suppose  the  wave  forms  of  the  two  currents 
had  been  identical  and  each  had  contained  a  third  harmonic.  A 
phase  displacement  of  60  degrees  between  the  waves,  i.e.,  between 
their  fundamentals,  would  have  made  the  third  harmonic  in  the 
resultant  zero.  The  effect  of  the  third  harmonics  would  not 
have  canceled  had  the  equivalent  sine  waves  been  added. 

It  is  readily  seen  from  what  precedes  that  whenever  the 
voltage,  current  and  power  in  the  component  parts  of  a  circuit 
with  parallel  branches  are  measured,  and  the  equivalent  sine 
waves  of  current  determined  from  the  instrument  readings  are 
added  vectorially  as  if  they  were  really  sinusoidal  waves,  the 
resultant  current  thus  determined  may  be  in  considerable  error 
if  the  wave  forms  of  the  component  currents  differ  greatly  from 
sinusoids.  A  similar  statement  holds  regarding  the  addition  of 
the  equivalent  sine  waves  of  voltage  drop  across  the  component 
parts  of  a  series  circuit. 


CHAPTER  V 

CIRCUITS  CONTAINING  RESISTANCE,  INDUCTANCE  AND 
CAPACITANCE 


Coefficient  of  Self-induction  or  the  Self-inductance  of  a 
Circuit.  —  In  the  neighborhood  of  an  electric  circuit  carrying  a 
current,  there  exists  a  magnetic  field  whose  intensity  at  any 
point  is  dependent  upon  the  strength  of  the  current,  the  con- 
figuration of  the  circuit  and  the  distance  of  the  point  from  the 
circuit.  If  the  current  alters  its  value,  the  field  is  also  altered, 
increasing  with  increase  of  current  and  decreasing  with  decrease 
of  current.  This  magnetic  field  is  a  definite  seat  of  energy  and 
for  its  production  requires,  therefore,  a  definite  expenditure  of 
energy,  determined  in  amount  by  the  flux  and  the  conducting 
ampere-turns  of  the  circuit  with  which  this  flux  is  linked. 

The  linkages  of  flux  with  turns  constitutes  one  of  the  most 
important  factors  of  any  circuit.  The  change  in  the  number  of 
linkages  per  unit  current  for  an  electric  circuit  is  called  the 
coefficient  of  self-induction  or  the  self-inductance.  The  self- 
inductance  of  the  circuit  is  denoted  by  the  symbol  L. 

If  d(p  is  the  change  in  flux  linking  a  circuit  of  N  turns  produced 
by  a  change  di  in  the  current,  the  coefficient  of  self-induction  or 
the  self-inductance  of  the  circuit  is 


It  is  the  rate  of  change  of  flux  linkages  of  a  circuit  with  respect 
to  the  current  it  carries.  If  the  flux  linkages  per  unit  current  are 
constant,  the  coefficient  of  self-induction  may  be  written 


where  <p  is  the  flux,  produced  by  the  current  /,  which  links  with 
the  N  turns.  In  general,  all  the  flux  does  not  link  with  all  the 
turns  of  the  circuit.  In  such  cases  the  calculation  of  the  coeffi- 
cient of  self-induction  becomes  more  or  less  difficult.  In  most 
cases  accurate  calculation  is  impossible. 

115 


116  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Coefficient  of  Mutual-induction  or  the  Mutual-inductance 
of  a  Circuit. — When  two  circuits  are  so  related  that  a  change  in 
the  current  in  one  produces  a  change  in  the  flux  linking  the  other, 
the  circuits  are  said  to  possess  mutual-inductance.  The  mutual- 
inductance  or  coefficient  of  mutual-induction  of  a  circuit  with 
respect  to  another  circuit  is  the  change  in  the  flux  linkages  of  the 
second  produced  by  a  change  of  one  unit  of  current  in  the  first. 
In  other  words,  it  is  the  rate  of  change  of  flux  linkages  of  the 
second  circuit  with  respect  to  the  current  in  the  first.  Mutual- 
inductance  will  be  considered  more  in  detail  later. 

Henry,  Secohm  or  Quadrant. — When  the  number  of  flux 
linkages  due  to  one  abampere  flowing  in  a  circuit  is  109,  the  circuit 
is  said  to  possess  a  self-inductance  of  one  henry.  This  definition 
holds  only  when  flux  is  proportional  to  current,  in  other  words 
it  is  strictly  true  only  when  there  is  no  magnetic  material  pres- 
ent. When  there  is  no  magnetic  material  present,  self-inductance 
is  constant.  When  flux  is  not  proportional  to  current,  a  circuit  is 
said  to  have  a  self-inductance  of  one  henry  when  the  rate  of 
change  of  flux  linkages  with  respect  to  current  in  abamperes  is 
109.  The  inductance  of  such  a  circuit  is  also  denoted  as  one 
quadrant,  for  the  reason  that  the  earth  quadrant  is  109  centi- 
meters. It  is  likewise  denoted  by  one  secohm,  an  abbreviation 
of  second-ohm  and  therefore  having  the  requisite  dimension, 
namely  length.  The  name  Henry  is  in  honor  of  Joseph  Henry, 
one  of  the  earliest  distinguished  workers  in  electromagnetic 
phenomena. 

Energy  of  the  Field. — If  the  current  in  a  circuit  remains  con- 
stant in  value,  there  is  no  expenditure  of  energy  in  maintaining 
the  field.  This  excludes  the  energy  dissipated  in  heat  in  the 
electric  circuit  itself  due  to  the  Pr  loss.  If,  however,  the  field 
increases,  there  will  be  a  reaction  developed  which  will  require 
an  expenditure  of  electrical  energy  by  the  circuit  to  overcome 
it.  This  electrical  energy  appears  as  the  magnetic  energy  of 
the  increased  field.  If,  on  the  other  hand,  the  field  diminishes, 
there  will  be  a  reaction  in  the  opposite  direction  and  in  virtue 
of  this,  energy  will  be  contributed  by  the  magnetic  field  to  the 
electric  circuit.  The  reaction  in  each  case  takes  the  form  of  an 
electromotive  force,  the  magnitude  of  which  depends  on  the 
time  rate  of  change  of  flux  linkages.  This  electromotive  force 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        117 

is  known  as  the  electromotive  force  of  self-induction.     Expressed 
as  a  rise  in  electromotive  force  it  is 


where  e  is  the  electromotive  force  of  self-induction,  N  the  number 
of  conducting  turns  in  the  circuit,  and  v  the  flux  linked  with  the 
turns  N.  N<p  is  the  number  of  flux  linkages  of  the  circuit. 

Effect  of  Self  -inductance  for  a  Circuit  Carrying  an  Alternating 
Current.  —  It  is  evident  that  if  a  circuit  carries  an  alternating 
current  such,  for  example,  as  a  simple  harmonic  current,  there  will 
be  an  alternate  increase  and  diminution  in  the  energy  of  the 
magnetic  field  and  this  will  give  rise  to  a  reactive  electromotive 
force.  If  a  complete  period  for  the  current  is  considered,  it  will 
be  found  that  during  one-half  of  this  period  energy  is  supplied  by 
the  circuit  to  the  magnetic  field  and  during  the  other  half  of  the 
period  energy  is  supplied  by  the  field  to  the  circuit.  When  the 
current  is  increasing  in  either  a  positive  or  a  negative  direction, 
the  establishment  of  energy  in  the  field  sets  up  an  opposing 
electromotive  force  which  retards  the  flow  of  current,  thus 
decreasing  its  rate  of  increase.  This  results  in  the  current  reach- 
ing a  given  value  later  than  it  would  have  reached  it  provided 
there  were  no  such  opposing  electromotive  force.  While  the 
current  is  decreasing,  the  field  contributes  energy  to  the  circuit 
and  diminishes  the  rate  at  which  the  current  falls,  thus  causing  it 
to  pass  through  a  given  value  later  than  it  would  have  done  pro- 
vided no  energy  were  returned  to  the  circuit  by  the  field.  The 
net  effect  is  to  "decrease  the  maximum  positive  and  negative 
values  reached  by  the  current  during  a  cycle  and  to  cause  the 
current  to  lag  behind  the  impressed  electromotive  force  producing 
it.  In  the  flow,  therefore,  of  a  simple  harmonic  current  in  a 
circuit  which  possesses  self  -inductance,  the  value  of  the  current 
will  be  less  than  if  there  were  no  self-inductance  and  the  current 
will  lag  by  a  certain  angle  with  respect  to  the  impressed  electro- 
motive force. 

Inductance  and  resistance  are  very  different  in  their  effects. 
Inductance  opposes  only  a  change  in  the  current  and  is  like  mass 
in  mechanics.  Resistance  opposes  the  flow  of  a  steady  current 
as  well  as  a  variable  current  and  its  analogue  is  friction  in  mech- 


118  PRINCIPLES  OF  ALTERNATING  CURRENTS 

anics.  The  kinetic  energy  of  a  moving  mass  is  ^MV2,  where  M 
is  the  mass  and  V  its  velocity.  For  the  electric  circuit  containing 
self-inductance,  the  kinetic  energy  is  Q^2>  where  L  is  the  self- 
inductance  and  /  the  current.  For  the  circuit,  the  kinetic  energy 
is  the  energy  in  the  magnetic  field  set  up  by  the  current. 

Capacitance. — The  capacitance  of  a  condenser  is  measured  by 
the  charge  required  to  raise  its  potential  by  unity.  It  is  equal 
to  the  ratio  of  charge  to  voltage. 

c-Q 

V 

A  condenser  is  said  to  have  a  capacitance  of  one  farad  when 
a  charge  of  one  coulomb  raises  it  to  a  potential  of  one  volt. 
This  unit  is  too  large  for  practical  use.  For  this  reason  the 
capacitance  of  condensers  is  ordinarily  expressed  in  microfarads. 
The  resistance  to  direct  current  of  a  well  made  condenser  is  very 
high  and  for  most  practical  purposes  may  be  considered  infinite. 
This  does  not  mean,  however,  that  a  condenser  connected 
across  an  alternating-current  circuit  will  take  no  current.  It 
will  alternately  charge  and  discharge  at  the  frequency  of  the 
circuit  in  which  it  is  connected  and  will,  therefore,  take  an 
alternating  current  of  perfectly  definite  effective  or  root-mean- 
square  value.  The  resistance  of  a  condenser  to  an  alternating 
current  is  equal  to  the  average  power  it  takes,  when  placed  across 
an  alternating-current  circuit,  divided  by  the  square  of  the 
effective  or  root-mean-square  current  taken  from  the  mains. 
The  power  absorbed  is  the  Pr  loss  in  the  condenser  caused  by  the 
alternating  charging  current,  plus  the  hysteresis  loss  in  the 
dielectric.  The  latter  loss  is  caused  by  the  varying  stresses 
produced  in  the  dielectric  by  the  alternating  voltage  impressed 
across  the  condenser  terminals.  The  dielectric  hysteresis  loss  per 
unit  volume  depends  on  the  nature  of  the  dielectric,  the  maxi- 
mum potential  gradient  to  which  it  is  subjected  and  the  fre- 
quency. The  dielectric  hysteresis  loss  for  air  is  zero.  It  is  small 
in  most  dielectrics  at  commercial  frequencies,  i.e.,  at  frequencies 
of  60  cycles  or  lower.  At  ordinary  frequencies  the  losses  in 
commercial  condensers  are  small  and  for  most  purposes  may 
be  neglected. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       119 

Effect  of  a  Condenser  in  an  Alternating-current  Circuit.  — 
The  effect  of  a  condenser  in  a  circuit,  so  far  as  the  phase  relation 
between  current  and  voltage  is  concerned,  is  just  the  opposite 
to  that  of  inductance.  Inductance  causes  the  current  in  a 
circuit  to  lag  the  voltage  drop  across  its  terminals.  A  condenser 
causes  the  current  in  a  circuit  to  lead  the  voltage  drop  across 
its  terminals. 

The    voltage    drop    across    the  terminals  of  a  condenser  is 


where  C  is  the  capacitance  of  the  condenser  and  q  its  charge. 
When  the  charge  on  the  condenser  is  a  maximum,  the  voltage 
drop  across  its  terminals  is  also  a  maximum.  When  the  charge 
is  zero,  the  voltage  drop  is  also  zero.  The  current  taken  by  a 
condenser  without  leakance  at  any  instant  is  equal  to  the  dif- 
ference between  the  voltage  impressed  across  its  terminals  and 
the  opposing  electromotive  force  due  to  the  condenser  charge, 
divided  by  the  condenser  resistance.  The  opposing  electro- 

motive force  is  e  =  ^.     Therefore,  when  the  charge  is  a  maxi- 
C 

mum  the  current  must  be  a  minimum  and  when  the  charge  is  a 
minimum  the  current  must  be  a  maximum.  There  can  be  no 
charge  in  the  condenser  until  there  has  been  current  flow,  since 
q  =  fidt.  The  current  must  consequently  lead  the  charge  and, 
since  the  voltage  drop  across  the  terminals  of  a  condenser  is 

e  =  ^,  it  must  also  lead  the  voltage  drop. 
C 

The  effect  of  a  condenser  in  an  electric  circuit  is  similar  to 
the  effect  of  elasticity  in  a  mechanical  system.  As  a  con- 
denser charges,  its  back  electromotive  force  rises  with  the 
charge  and  offers  an  increasing  opposition  to  further  current 
flow,  the  opposition  increasing  in  proportion  to  the  charge.  This 
back  electromotive  force  is  similar  in  its  effect  on  current  flow 
to  the  reaction  of  a  spring,  which  is  being  stretched.  The 
reaction  of  the  spring  offers  increasing  opposition  to  further 
stretching. 

Circuit  Containing  Constant  Resistance  and  Constant  Self- 
inductance  in  Series.  —  In  any  energy  relation  whatsoever,  the  sum 
of  the  actions  and  the  reactions  must  be  zero.  The  action  for  the 


120  PRINCIPLES  OF  ALTERNATING  CURRENTS 

ordinary  electric  circuit  is  balanced  by  the  impressed  voltage 
drop.  The  reactions  for  a  circuit  containing  resistance  and 
self-inductance  in  series  are  the  ohmic  and  reactive  drops,  the 
latter  being  due  to  the  presence  of  self-induction.  The  condition 
of  the  circuit  is  completely  determined  by  the  equation 


In  general  this  can  be  solved  only  when  <p  is  proportional 
to  i,  that  is,  when  L  is  constant.  All  voltages  in  equation  (1)  are 

di 
voltage  drops.     The  voltage  rise  due  to  self-induction  is  —L-r- 

The  drop  is  therefore  -\-L~rr. 

The  voltage  e  in  equation  (1)  is  the  voltage  drop  across  the 
circuit.  The  component  of  this  voltage  to  balance  the  ohmic 

di 
reaction  is  ri.    L-ris  the  component  to  balance  the  reaction  due 

to  self-induction,  i.e.,  the  electromotive  force  of  self-induction. 
If  instead  of  considering  reactions,  the  energy  relations  are 
considered,  the  following  evidently  holds: 

eidt  =  ri*dt+L^idt  (2) 

at 

Consider  the  total  energy  concerned  in  a  time  T,  equal  to 
the  periodic  time  for  the  circuit  in  question.  This  is  given  in 
the  following  equation: 


fT  f*T  CT      d' 

ei  dt  =        ri*dt  +    \     L  ~  idt 
Jo  Jo  Jo      dt 


(3) 


The  first  term  of  the  second  member  of  equation  (3)  is  the 
energy  dissipated  as  heat.  The  second  term  represents  the 
energy  stored  in  the  magnetic  field.  It  is  clear  that  if  the  self- 
inductance  of  the  circuit  is  constant,  this  second  term  becomes 
zero  when  integrated  over  a  complete  cycle,  as  would  be  expected, 
since  the  amount  of  energy  delivered  to  the  field  by  the  circuit 
during  one-half  of  any  complete  period  is  balanced  by  the  amount 
of  energy  delivered  to  the  circuit  by  the  field  during  the  other 
half  period. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       121 

Solution  of  the  Differential  Equation  of  a  Circuit  Containing 
Constant  Resistance  and  Constant  Self-inductance  in  Series.— 

The  solution  of  the  equation 


giving  the  current  in  terms  of  e,  r,  L, 
and  t  will  evidently  depend  for  its 
form  on  e. 

Case  I. — Growth  of  Current  in  an 
Inductive  Circuit  on  which  a  Constant 
Electromotive  Force  is  Impressed. —  . 

(Fig.  33,  key  on  a.)  FIG.  33. 

e  =  E  =  constant. 
E=ri+L^  (5) 

This  is  a  linear  differential  equation  of  the  first  order  with 
constant  coefficients.  The  general  solution  of  such  an  equation 
is  of  the  form 

i  =  Y  +  u  (6) 

where  Y  is  the  complementary  function  and  u  is  the  particular  in- 
tegral. (Murray,  Differential  Equations,  page  63,  second  edition.) 
In  the  case  of  an  electric  circuit,  Y  is  the  transient  term  in  the 
expression  for  the  current.  This  term  persists  only  during  the 
establishment  of  the  current  and  becomes  zero  when  steady  condi- 
tions have  been  reached.  The  particular  integral  represents  the 
steady  state,  i.e.,  the  condition  in  the  circuit  after  the  transient 
term  has  become  zero.  The  general  equation  for  the  establish- 
ment of  a  current  in  an  electric  circuit  always  contains  a  transient 
and  a  steady  term.  The  conditions  in  the  circuit  at  any  instant 
are  represented  by  the  sum  of  the  two  terms.  Usually  the 
transient  term  is  sensibly  equal  to  zero  after  a  short  interval  of 
time. 

Since  the  impressed  voltage  E  in  equation  (5)  is  constant,  the 
current  must  follow  Ohm's  law  after  steady  conditions  have  been 
established.  The  particular  integral,  u,  therefore  must  be  equal 

E 

to  — . 


122  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  complementary  function  is  always  found  by  making  the 
impressed  voltage  zero,  in  this  case  by  making  E  zero.  The 
solution  for  the  current  is  of  the  form  i  =  Aeat.  (Murray, 
Differential  Equations,  page  63.  second  edition.)  Substituting 
E  =  0  and  i  =  Aeat  in  equation  (5)  gives 

0  =  rAeat  +  aALeat  (7) 

This  holds  for  all  values  of  t. 

0  =  r  +  aL  (8) 

Equation  (8)  is  an  equation  of  the  first  degree  and  therefore 

has  but  a  single  root. 

r 
a=    ~L 

The  complete  solution  of  equation  (5)  is  therefore 

_rt        p 

1  =  Ae     r+  -  (9) 

where  A  is  a  constant  of  integration  which  must  be  determined 
from  the  conditions  existing  in  the  circuit  at  the  instant  t  =  0. 
When  t  is  zero  i  in  most  cases  is  also  zero.  Putting  both  t  and  i 
equal  to  zero  gives 

A  =  _E 
r 
Therefore 


At  the  instant  of  closing  the  circuit  t  is  zero  and  the  current 

di 
i  is  also  zero.     The  rate  of  change  of  current,  T.,  is  a  maximum, 

being  equal  to  -j-     (Equation  (5))     As  t  increases,  -^  decreases, 

approaching  zero  as  a  limit,  while  the  current  approaches  the 
limiting  or  Ohm's  law  value. 

If  the  conditions  in  the  circuit  are  such  that  the  current 
is  not  zero  when  t  is  zero,  i.e.,  at  the  instant  the  electromotive 
force  E  is  applied  to  the  circuit,  the  constant  A  may  still  be 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       123 

evaluated.  Let  70  be  the  current  in  the  circuit  at  the  instant  E 
is  added  and  let  EQ  be  the  electromotive  force  producing  this 
current.  Then  from  equation  (9)  when  t  =  0 


Therefore  in  this  case 


If  the  current  70  has  reached  its  steady  value,  i.e.,  its  Ohm's 

E 
law  value  — ,  when  E  is  added,  equation  (12)  becomes 


Equation  (13)  reduces  immediately  to  equation  (11)  when  70 
and  EQ  are  zero,  i.e.,  when  there  is  no  current  or  voltage  in  the 
circuit  when  E  is  impressed. 

It  cannot  be  emphasized  too  strongly  that  the  expression 

for  the  current  is  always  made  up  of  two  terms.     One,  — ,  equa- 

]?    I   p1 

tion  (10),  and  -  — ,  equation  (12),  gives  the  value  of  the  cur- 
rent after  steady  conditions  have  been  reached.  The  other, 

_rt  _rt 

-«   L,  equation  (10),  and  (70 )  €      ,  equation  (12),  is 

a  transient  which  decreases  logarithmically  and  becomes  zero 
when  steady  conditions  have  been  attained.  The  coefficient  of 
the  transient  term  is  determined  by  the  conditions  existing  in 
the  circuit  at  the  instant  the  electromotive  force  is  impressed  and 
by  the  steady  state.  See  equation  (12). 

Theoretically  it  takes  an  infinite  time  for  the  transient  term 
to  become  zero,  but  in  practice  it  usually  becomes  negligibly  small 
in  a  comparatively  brief  interval  of  time. 

The  rate  of  increase  of  current  depends  on  the  ratio  of  the 
self-inductance  to  the  resistance,  i.e.,  on  the  ratio  of  L  to  r. 
Either  large  resistance  or  small  inductance  will  make  the  rate  of 


124 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


increase  of  current  rapid.     As  this  rate  of  increase  is  determined 
by  the  ratio  of  L  to  r,  —  is  known  as  the  time  constant  of  the 


-  represents  the  time  required  for  the  current  to  reach 


circuit. 

0.632  of  its  final  or  Ohm's  law  value. 

Steady  and  transient  terms  similar  to  those  in  equations  (10) 

and  (12)  are  typical  and  occur  in  all  equations  for  current  in 

circuits  containing  resist- 
ance and  inductance,  re- 
sistance and  capacitance, 
or  resistance,  inductance 
and  capacitance,  in  which 
the  current  has  not  reached 
its  steady  state.  The 
growth  of  current  in  a 
circuit  containing  resist- 
ance and  self -inductance  in 
series  is  shown  graphically 

in  Fig.  34.     In  this  figure  the  current  is  assumed  to  be  zero 

when  t  is  zero. 

All  three  curves  shown  are  for  the  same  impressed  voltage, 

E,  and  for  the  same  resistance,  r.    The  self -inductance,  L,  is 

least  in  curve  a  and  greatest  in  curve  c. 
Energy  in    the    Electromagnetic    Field. — From    the    energy 

equation  (equation  (3),  page  120)  it  will  be  seen  that  energy  is 

being  constantly  dissipated  in  heat  and  also  is  being  stored  in  the 

field  at  a  rate,  L -ri,  which  constantly  decreases.    The  total  energy 

thus  stored  in  the  magnetic  field  up  to  a  time  t,  at  which  the 
current  is  /,  is  evidently  given  by 


Time 
FIG.  34. 


^idt=       Lidi  =  ~LP 


(14) 


1 


The  expression  ~  LP  is  sometimes  called  the  electrokinetic 

energy  of  the  circuit.  It  is  analogous  to  the  expression  \MVZ 
for  the  kinetic  energy  of  a  moving  body,  where  M  is  the  mass  of 
the  body  and  V  is  its  velocity. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       125 

The  expression 

W=\LI*  (15) 

for  the  electrokinetic  energy  of  a  circuit  holds  only  when  the 
self-inductance,  L,  is  constant.  If  the  circuit  contains  magnetic 
material,  L  is  not  constant,  but  is  a  function  of  the  current  in 
the  circuit.  If  L  =  f(i), 

W  =  \t(i)idi  (16) 

Jo 

The  flux  density  in  a  magnetic  circuit  containing  iron  may  be 
expressed,  approximately,  in  terms  of  the  current  by  the  follow- 
ing empirical  equation, 

' 


where  the  k's  are  constants  which  may  be  obtained  readily  from  a 
magnetization  curve.  (B  is  the  flux  density  and  N  is  the  number 
of  turns  in  the  winding  on  the  magnetic  circuit.  If  A  is  the 
cross  section  of  the  magnetic  circuit 

L  =  NA^  (18) 


Equations  (17)  and  (18)  assume  there  is  no  leakage  of  flux 
between  turns,  that  is,  that  all  the  flux  produced  by  any  one  turn 
of  the  winding  links  all  of  the  turns.  This  condition  is  never 
exactly  fulfilled  in  practice,  although  in  some  cases  it  is  approxi- 
mately attained. 

Case  II.  —  Decay  of  Current  in  an  Inductive  Circuit  when  the 
Impressed  Electromotive  Force  is  Removed  by  Short  Circuiting.  — 
(Fig.  33,  page  121,  key  on  b.)  The  switch  is  assumed  to  be  so  ar- 
ranged as  not  to  break  the  circuit  when  it  is  thrown  from  a  to  b. 

When  E  is  zero,  equation  (4),  page  121,  becomes 

0  =  ri+Lg|  (19) 

The  solution  of  this  equation,  as  in  Case  7,  is  of  the  form 

i  =  Y  +  u 

but  now  the  particular  integral,  u,  which  represents  the  steady 
state,  is  zero.  The  complete  solution  of  equation  (19)  therefore 
contains  only  the  transient  term  and  is 

.  A.'*  <«» 


126 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


where,  as  before,  A  is  the  constant  of  integration  which  may  be 
evaluated  by  putting  i  equal  to  zero.  Assume  the  current  to 
have  attained  a  value  /</  when  the  electromotive  force  is  removed. 
Then  when  t  =  0,  i  =  !</  and 

A  =  i  =  I0' 

A  is  equal  to  the  value  of  the  current  in  the  circuit  at  the 
instant  the  electromotive  force  is  removed.  Putting  the  value  of 
A  in  equation  (20)  gives 

(21) 


0  € 


If  the  current  has  reached  its  Ohm's  law  value  before  the 
electromotive  force  is  removed,  equation  (21)  becomes 


From  the  energy  equation  (equation  (2),  page  120). 
0 


ri*dt  +  L^.idt 
at 


(22) 


(23) 


=  10  ohms 


or 


ri*dt  =  —Lidi 


Time          2     Seconds 

FIG.  35. 


The  energy  in  the  field 
is  gradually  being  dis- 
sipated in  heat  in  the 
circuit. 

The     conditions     repre- 
sented   by    equation    (22) 
are  shown   graphically   in 
Fig.  35. 
The  rate  of  decay  of  the  current  depends  on  the  ratio  of 

self -inductance  to  resistance,  i.e.,  on  the  time  constant  —  .   Either 

r 

small  self-inductance  or  large  resistance  will  make  the  decay 
of  current  rapid.  The  time  constant  represents  the  time  re- 
quired for  the  current  in  a  circuit  to  reach  0.632  of  its  Ohm's 
law  value  when  the  circuit  is  closed,  or  to  fall  to  (1  —  0.632)  = 
0.368  of  its  Ohm's  law  value  when  the  electromotive  force  is 
removed.  The  above  statements  regarding  the  time  constant 
assume  that  the  current  is  zero  when  the  circuit  is  closed  and  has 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       127 

reached  its  Ohm's  law  value  when  the  electromotive  force  is 
removed. 

Breaking  an  Inductive  Circuit. — If  instead  of  removing  the 
electromotive  force,  i.e.,  short-circuiting  it,  as  in  Case  II,  the 
circuit  is  broken,  the  decay  of  current  will  be  much  more  rapid, 
owing  to  the  great  increase  in  resistance  caused  by  the  introduc- 
tion of  the  air  gap  at  the  break.  The  energy  of  the  magnetic 
field  must  be  delivered  to  the  circuit  in  the  brief  interval  of  time 
required  to  interrupt  the  current.  The  electromotive  force 
induced  when  the  circuit  is  broken  may  be  very  great,  and  indeed 
may  be  sufficiently  high  to  establish  an  arc  across  the  terminals 
of  the  circuit,  even  though  the  electromotive  force  impressed 
initially  on  the  circuit  to  produce  the  current  would  not  be  suffi- 
cient to  start  the  arc.  If  it  were  possible  to  break  a  circuit  in 
zero  time,  without  the  dissipation  of  any  energy  at  the  break, 
the  self-induced  voltage  would  be  infinite.  In  this  case  there 
would  be  infinite  voltage  across  the  break.  Such  a  condition 
is  not  possible  in  practice,  but  very  high  voltages  will  be  produced 
whenever  inductive  circuits  are  broken  rapidly.  Special  pre- 
cautions must  be  taken  when  highly  inductive  circuits,  such  as 
the  fields  of  motors  and  generators,  are  opened. 

If  an  inductive  circuit  is  shunted  by  a  suitable  non-inductive 
resistance  before  being  disconnected  from  the  mains,  the  rise  in 
voltage  across  its  terminals  may  be  limited  to  any  desired  amount. 
If  the  shunt  could  be  made  absolutely  non-inductive  and  its 
resistance  were  equal  to  the  resistance  of  the  inductive  circuit  to 
be  broken,  no  rise  in  voltage  could  occur  across  the  terminals  of 
the  inductive  circuit  even  if  the  switch  disconnecting  it  with  its 
shunt  from  the  mains  could  be  opened  in  zero  time.  If  the 
switch  could  be  opened  in  zero  time,  the  current  in  the  inductive 
circuit  would  be  immediately  established  in  the  shunt  and  would 
then  decrease  at  a  rate  determined  by  equation  (21),  page  126, 
where  r  is  the  resistance  of  the  shunt  plus  the  resistance  of  the 
inductive  circuit.  The  highest  possible  voltage  across  the 
inductive  circuit  before  it  is  opened  is  Ir  =  E  where  r  is  its 
resistance  and  E  is  the  voltage  of  the  circuit  to  which  it  is  con- 
nected. This  voltage  would  be  reached  if  the  current  had 

ET 

attained  its  Ohm's  law  value,  /  =  — ,  before  breaking  the  circuit. 


128  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Under  this  condition,  the  current  in  the  shunt  the  instant  after 

E 
opening  the  circuit  would  be  I  —  —.    The  voltage  across  its 

terminals  would  be  Ira,  where  rs  is  the  resistance  of  the  shunt. 
This  voltage  would  be  equal  to  E  if  ra  were  made  equal  to  r,  the 
resistance  of  the  inductive  circuit.  When  the  field  circuits 
of  large  motors  or  generators  have  to  be  disconnected  from  the 
mains  while  carrying  current,  it  is  customary  to  provide  them 
with  suitable  non-inductive  shunts  to  prevent  a  dangerous  rise 
in  voltage.  Such  shunts  are  called  field  discharge  resistances. 
They  are  used  with  special  field  discharge  switches  which  con- 
nect them  in  circuit  only  while  the  current  is  being  interrupted. 
Cose  III. — Simple  Harmonic  Electromotive  Force  Impressed 
on  a  Circuit  Containing  Constant  Resistance  and  Constant  Self- 
inductance  in  Series. — In  this  case,  e  =  Em  sin  (ut  +  a)  and 
equation  (4),  page  121,  becomes 

Em  sin  (ut  +  a)  =  ri  +  L^  (24) 

This  is  a  linear  differential  equation  of  the  first  order  with 
constant  coefficients.  Its  first  term  is  a  function  of  t.  The  solu- 
tion of  the  equation  is  again  of  the  form 

i  =  Y  +  u 

The  transient  term,  F,  is  found  as  in  Cases  /  and  //.  From 
equation  (9)  the  transient  is 

Y  =  Ae~£  (25) 

The  constant  A  is  determined  by  the  conditions  in  the  circuit 
at  the  instant  t  =  0. 

The  particular  integral  may  be  evaluated  most  easily  in  the 
following  manner.  Under  steady  conditions,  the  drop  in  voltage 
across  the  circuit  will  be  made  up  of  two  parts :  one,  ri,  due  to  the 

resistance,  the  other,  L-r,  due  to  the  self-inductance.     If  the 

resistance  is  constant,  the  drop  caused  by  it  will  be  of  the  same 
wave  form  as  the  current.  The  drop  caused  by  the  self-induc- 
tance, L,  will  not,  however,  be  of  the  same  wave  form  as  the 
current,  even  when  L  is  constant,  except  when  the  current  is 
sinusoidal.  If  the  current  is  sinusoidal  and  L  is  constant,  the 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       129 

drop  L  ,    will  also  be  sinusoidal,  since  the  derivative  of  the 

current  with  respect  to  time  will  then  be  a  cosine  function  of 
time  which  is  equivalent  to  a  sine  function  advanced  90  degrees 
in  phase.  Since  the  sum  of  any  number  of  sinusoidal  waves  of 
the  same  frequency  is  also  a  sinusoidal  wave  of  like  frequency, 
it  follows  that  if  r  and  L  are  both  constant  and  the  current  is 
sinusoidal,  the  voltage  drop  impressed  on  the  circuit  must  also 
be  sinusoidal,  since  it  is  the  sum  of  the  two  sinusoidal  drops,  in 
quadrature  with  each  other,  due  to  the  current.  Conversely,  if 
the  voltage  impressed  on  the  circuit  is  sinusoidal,  the  sum  of 
the  reactions  due  to  the  current  must  be  sinusoidal.  The  only 
way  the  sum  of  these  reactions  can  be  sinusoidal  when  r  and  L 
are  constant  is  for  the  current  to  be  sinusoidal. 
Let  the  current  be 

i  =  Im  sin  (o>*  +  a  +  0)  (26) 

where  Im  is  the  maximum  value  of  the  current  after  steady  con- 
ditions have  been  attained  and  6  is  its  phase  angle  with  respect 
to  the  voltage  Em. 

Substituting  the  value  of  the  current  given  by  equation  (26)  in 
equation  (24)  gives 

Em  sin  (wt  +  a)  =  rlm  sin  (cot  +  a  +  9) 


=  rlm  sin  (ut  +  a  +  0) 

-f  coL/m  cos  M  +  a  +  B)          (27) 

Since  a  cosine  function  of  time  leads  a  sine  function  by  90 
degrees,  equation  (27)  may  be  written  in  the  following  form 

Em  sin  (ut  +  a)  —  rlm  sin  (co£  +  a  +  0) 

sin  (««  +  a  +  0  +  90°)       (28) 


The  two  terms  in  the  right-hand  member  of  equation  (28)  are 
two  quadrature  components  of  the  voltage  drop,  Em  sin  (ut  +  «), 
across  the  circuit.  The  maximum  values  of  these  components 
are  rlm  and  coL/m.  The  component  wL7m  leads  the  component 
rlm  by  90  degrees.  These  two  components  are  respectively  the 
reactive  and  the  active  components  of  the  voltage  drop. 


130 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


The  vectors  corresponding  to  the  three  terms  of  equation  (28) 

a  +  0 
are  plotted  in  Fig.  36  for  the  instant  of  time  t  =    —  —    — .     At 


CO 


Fia.  36. 


this  instant  the  vector  represent- 
ing the  current  lies  along  the 
axis  of  reals. 

The  waves  corresponding  to  the 
vectors  are  plotted  in  Fig.  37,  0  be- 
ing assumed  greater  than  a. 

Referring  to  Fig.  36  it  is  obvious 
that 

=  /mVrM7^  (29) 


and 


FIG.  37. 

The  voltage  Em  leads  the  current  Im  by  an  angle  0  or  the  current 
lags  the  voltage  by  the  same  angle.  Since  0  in  equations  (27)  and 
(28)  is  the  phase  angle  of  the  current  with  respect  to  the  voltage, 
it  should  be  considered  negative  as  the  current  lags  the  voltage. 

The  particular  integral  or  steady  term  in  the  general  equation 
for  the  current  in  a  circuit  having  constant  resistance  and  con- 
stant self-inductance  in  series  is  consequently 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        131 

(32) 
The  complete  solution  of  equation  (24)  is  therefore 

rt  Jf  /  j\ 

i  =  A€~L  -j m       =  sin  tut  -f-  a  —  tan"1—  )      (33) 

V**2  +  co2L2         \  r  / 

When  t  is  zero  the  current  i  is  also  zero.     Putting  both  t 
and  i  equal  to  zero  in  equation  (33)  gives 


-Em 


in  [  a  —  tan  l 


Therefore 

-Bt 


\/r2+a>2L2 


sin 


(«  -  taar'— 


I.  +Vr^T^Sin("<  +  a~ta°"'T)     (34) 

UMJ 

The  angle   6  =  tan"1  —  is  the  angle  of  lag  of  the  current 

behind  the  voltage  after  steady  conditions  have  been  attained. 
It  is  determined  by  the  constants  of  the  circuit  and  theoretically 
at  least  may  have  any  value  between  0  and  90  degrees.  It  is  zero 
when  the  inductance  is  zero,  i.e.,  for  a  non-inductive  circuit. 
It  would  be  90  degrees  for  a  circuit  having  inductance  but  no 
resistance,  if  such  a  condition  were  possible  of  attainment.  In 
practice,  6  may  be  very  nearly  90  degrees  but  it  can  never  be 
exactly  90  degrees,  since  it  is  impossible  to  have  a  circuit  without 
some  resistance.  There  is  no  difficulty  in  making  a  circuit  prac- 
tically non-inductive  for  ordinary  frequencies. 

The  magnitude  of  the  transient  term  in  the  expression  for  the 
current  (first  term  in  the  second  member  of  equation  (34))  is  deter- 
mined by  the  particular  point  on  the  electromotive  force  wave  at 
which  the  circuit  is  closed.  The  magnitude  of  the  phase  angle  a, 
in  the  expression  for  the  instantaneous  voltage,  equation  (24), 
page  128,  fixes  the  value  of  the  voltage  e  when  t  is  zero.  The 

7T 

transient  will  be  a  maximum  when  (a  —  6)  =  9  or  wnen  the 
circuit  is  closed  at  that  point  on  the  electromotive  force 


132 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


wave  which  corresponds  to  maximum  current  after  steady  con- 
ditions have  been  established.  The  transient  will  be  zero  when 
(a  _  0)  =  o,  i.e.,  when  the  circuit  is  closed  at  that  point  on  the 
electromotive  force  wave  which  corresponds  to  zero  current  after 
steady  conditions  have  been  reached. 

Fig.  38  shows  the  resultant  current  and  its  transient  and 
steady  components  when  a  circuit  containing  constant  resistance 
and  constant  self -inductance  in  series  is  closed  at  the  point  on  the 
electromotive  force  wave  which  makes  the  transient  a  maximum. 
The  figure  is  for  a  25-cycle  circuit  having  a  ratio  of  coL  to  r  of  15. 

Impressed  Voltage  shown  by  line  of  round  dots. 

Current  shown  by  full  line. 

Transient  and  steady  components  of  current  shown  by  lines  of  short  dashes. 


7"  =  0.0955.  Frequency  ~  25  cycles,  -f-  =  15,  Angle  of  lag  •=  86.2  degrees. 
Curves  show  the  conditions  existing  when  the  circuit  is  closed  at  the  point 
on  the  voltage  wave  which  makes  the  transient  a  maximum,  a-  6  -  -|- 

FlG.  38. 


It  is  obvious  from  Fig.  38  that  the  maximum  value  the  current 
can  attain,  when  a  circuit  having  constant  resistance  and  constant 
self -inductance  in  series  is  closed,  can  never  be  equal  to  twice  the 
maximum  value  of  the  current  under  steady  conditions.  It  may 
reach  nearly  twice  that  value  in  a  high-frequency  circuit  having  a 
large  time  constant,  i.e.,  having  a  large  ratio  of  self-inductance 
to  resistance,  for  under  such  conditions  the  transient  will  have 
diminished  but  little  at  the  end  of  the  first  half  cycle  and  will 
then  add  to  the  steady  component  to  give  a  maximum  nearly 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       133 

equal  to  twice  the  maximum  value  of  the  current  under  steady 
conditions. 

Under  ordinary  conditions  the  transient  is  of  little  importance 
when  circuits  having  constant  resistance  and  constant  self- 
inductance  in  series  are  closed,  since  it  practically  disappears  after 
a  few  cycles  and  in  general  produces  no  dangerous  rise  in  current. 
In  the  case  plotted  in  Fig.  38,  the  transient  has  relatively  little 
effect  after  two  and  a  half  or  three  cycles,  that  is  after  about  a 
tenth  of  a  second,  and  this  is  for  a  circuit  having  a  relatively 
large  ratio  of  L  to  r. 

The  effect  of  the  transient  is  to  displace  the  axis  of  the  current 
wave  so  that  it  lies  along  a  logarithmic  curve,  which  is  the 
transient,  instead  of  lying  along  the  axis  of  time. 

Although  the  transient  is  of  relatively  little  importance  in 
most  cases,  it  is  of  importance  in  switching  operations  on  high- 
voltage  transmission  lines.  It  is  of  especial  importance  in  certain 
cases  where  the  inductance  is  not  constant  as,  for  example,  when 
an  alternator  is  short-circuited  or  when  a  large  transformer  is 
switched  on  a  line.  When  an  alternator  is  short-circuited,  the 
initial  maximum  value  of  the  short-circuit  current  may  reach  ten 
or  more  times  the  maximum  value  of  the  sustained  short-circuit 
current,  which  itself  may  be  several  times  the  rated  full-load 
current.  This  large  transient  short-circuit  current  is  caused,  in 
the  case  of  the  alternator,  by  a  very  great  decrease  in  the  apparent 
inductance  of  the  armature  winding  during  the  transient  period 
of  the  short  circuit. 

The  steady  current  for  a  circuit  having  constant  resistance  and 
constant  self-inductance  in  series  is  given  by 


Em 

i  =  — -====  sin 
A/r2  +  co2L2 


(  cot  +  a  —  tan"1  —  ] 
\  r  / 

=  Im  sin  (««  +  a  -  6)  (35) 

The  maximum  value  of  the  current  is  found  by  dividing  the 
maximum  value  of  the  voltage  by  \A2  +  co2L2.  It  lags  the 
maximum  value  of  the  voltage  by  an  angle  6  whose  tangent  is 
equal  to  the  ratio  of  coL  to  r.  Since  the  effective  value  of  a 
sinusoidal  current  is  equal  to  its  maximum  value  divided  by  the 
square  root  of  two,  it  is  evident  that  the  effective  value  of  the 
current  is  given  by  the  effective  value  of  the  voltage  divided  by 


134  PRINCIPLES  OF  ALTERNATING  CURRENTS 


\/r2  +  o>2L2.  The  quantity  V  r2  +  cu2L2  plays  a  similar  part  in 
alternating-current  circuits  to  resistance  in  direct-current  cir- 
cuits. If  it  is  denoted  by  the  letter  z,  the  expression  for  current 
may  be  written  in  a  form  similar  to  Ohm's  law  for  direct  cur- 
rents, i.e., 

T  E  E 

I  =  —  ,  =  =  — 

Vr2  +  co2L2        z 
Impedance  and  Reactance.  —  The  quantity 


z  =       r2  +  o>2L2 

is  called  the  impedance  and  is  measured  in  ohms.  It  is  a  constant 
only  when  the  resistance,  self-inductance  and  frequency  are 
constant.  The  expression  coL  is  called  the  inductive  reactance 
and  is  also  measured  in  ohms.  Reactance  is  denoted  by  the  letter 
x.  Thus  wL  =  x  and 


z  =       r  (37) 

Inductive  reactance  is  constant  only  when  frequency  and  self- 
inductance  are  constant.  It  is  the  inductive  reactance  of  a  cir- 
cuit which  causes  the  current  to  lag  in  phase  behind  the  voltage 
and  thus  to  have  a  quadrature  component.  It  is  because  this 
component  is  caused  by  reactance  that  it  is  called  the  reactive 
component. 

Vector  Method  of  Determining  the  Steady  Current  for  a  Circuit 
Having  Constant  Resistance  and  Constant  Self-inductance  in 
Series.  —  The  current  in  amperes  will  be  taken  along  the  axis  of 
reals.  Refer  to  Fig.  36,  page  130.  For  the  present  purpose  read 
Im  and  Em  on  the  figure  as  7  and  E  respectively,  i.e.,  as  root-mean- 
square  or  effective  values  instead  of  maximum  values.  The 
active  and  reactive  components  of  the  impressed  electromotive 
force  are  then  given  by  rl  and  juLI  =  jxl  respectively. 


E  =  rl  -{-jxl  =  I(r  +jz)  =    z 

In  complex,  therefore,  the  impedance  of  an  inductive  circuit  is 
given  by 

z  =  r+jx  (38) 

The  value  of  z  in  ohms  is  \A*2  +  #2-     Calling  the  component 
of  E  along  the  axis  of  reals,  i.e.,  the  real  or  active  component, 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       135 

Ea  and  the  component  along  the  axis  of  imaginaries,  i.e.,  the 
imaginary  or  reactive  component,  Er 

E  =     Ea  +  jEr 
E  (volts)  =  VEa2+  Er2 


Rationalizing  equation  (39),  to  get  rid  of  j  in  the  denominator, 
by  multiplying  the  numerator  and  the  denominator  by  the 
denominator  with  the  sign  of  the  term  involving  j  reversed,  gives 

j  =  Ea  +  jEr       r-jx 
r  +jx         r  -  jx 


Err  -Eax 

*         *    =/a  +  Jlr 


where 


Ear  +  Erx  Err  -  Eax 

and  Ir  -  -- 


are  respectively  the  active  and  reactive  components  of  the 
current.  But  /  is  along  the  axis  of  reals,  therefore  Ir  =  0. 
Hence  Err  =  Eax  and 

•'  ' 


where  6  is  the  tangent  of  the  angle  of  lag  of  the  current  behind  the 
voltage  E.     The  power  is 

Eala    =    El  COS  0.  (42) 

It  should  be  noted  that  impedance  is  not  a  vector  but  a  com- 
plex quantity.  When  it  is  multiplied  by  vector  current  vector 
voltage  results.  Complex  quantities  have  no  reference  axis  in 
the  ordinary  sense  but  when  they  are  multiplied  or  divided 
by  a  vector,  such  as  A,  the  result  is  a  new  vector  referred  to 
the  same  axis  as  the  vector  A  ,  but  usually  not  in  phase  with  A  . 
Although  complex  quantities,  such  as  impedance,  are  not  vectors, 
they  have  real  and  imaginary  parts  and  when  multiplied  or 
divided,  added  or  subtracted  must  be  treated  as  vectors  in  so 
far  as  these  operations  are  concerned, 


136  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Polar  Expression  for  the  Impedance  of  a  Circuit  Containing 
Constant  Resistance  and  Constant  Self-inductance  in  Series.— 
Multiplying  and  dividing  the  complex  expression  for  impedance 
by  \/r2  H-  x2  will  not  alter  its  value. 


=  0  +  jx) 


Vr*  +  x"  {  77===,  + 


=  z  (cos  0  +  j  sin  0)  (43) 

where  2  is  the  magnitude  of  the  impedance  and  6  is  its  angle, 

x 
i.e.,  the  angle  which  is  determined  by  the  relation  0  =  tan"1  — ' 

From  equation  (43)  it  is  seen  that  impedance  is  a  scalar 
quantity  multiplied  by  the  operator  (cos  0  +  j  sin  6)  which 
rotates  through  the  angle  0. 

Multiplying  current  expressed  in  complex  by  impedance,  also 
expressed  in  complex,  therefore  gives  the  correct  value  of  the 
impedance  drop  rotated  into  the  correct  phase  position  with 
respect  to  the  current. 

From  equation  (43)  it  is  obvious  that  the  polar  expression 
for  impedance  is 

z  =  z\0  (44) 

1" 

where    B  =  tan"1  -.     For    inductive    impedance    6    is    positive. 

Circuit  Containing  Constant  Resistance  and  Constant  Capaci- 
tance in  Series. — The  condition  of  a  circuit  containing  constant 
resistance  and  constant  capacitance  in  series  is  completely  deter- 
mined by  the  equation 

e=ri  +  %  (45) 

where  e  is  the  impressed  voltage  drop.     The  component  of  this 
to  supply  the  ohmic  drop  is  ri.     ^  is  the  component  to  overcome 

the  counter  electromotive  force  of  the  condenser.      If  q  is  the 
charge  on  the  condenser  in  coulombs  and  C  is  the  capacitance  of 

the  condenser  in  farads,  ^  is  the  voltage  drop  across  the  condenser 
in  volts. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        137 


The  energy  relation  is  shown  by 

eidt  =  ri2dt  +  ^ 


(46) 


Consider  the  total  energy  concerned  in  any  given  time    T. 

CT  CT  CTo 

eidt  =      ri2dt  +       fat  (47) 

Jo  Jo  Jo  L 

The  first  term  of  the  second  member  is  the  energy  dissipated 
in  joule  heating,  i.e.,  in  the  i2r  loss.  The  second  term  represents 
the  energy  stored  in  the  electrostatic  field  as  potential  energy 
due  to  the  strain  in  the  dielectric  of  the  condenser. 

Solution  of  the  Differential  Equation  for  a  Circuit  Containing 
Constant  Resistance  and  Constant  Capacitance  in  Series. — The 
solution  of  equation  (45)  giving  the  current  and  charge  of  the 
condenser  in  terms  of  e,  r,  C 
and  t,  will  evidently  depend  for 
its  form  on  the  impressed  electro- 
motive force. 

Case  I. — Growth  of  Charge  and 
Current  in  a  Capacitive  Circuit 
on  which  a  Constant  Electromotive 
Force  is  Impressed. — (Fig.  39,  key 
on  a.) 


UMAA/W 


FIG.  39. 


e  =  E  =  constant. 
E  =  ri  +  % 


Since 


E  =  r  ^  -f  *  (48) 

This  is  a  linear  differential  equation  of  the  first  order.  It  is  of 
the  same  form  as  equation  (5),  page  121  and  its  solution  is  of  the 
same  type. 

q  =  Y  +  u 

Obviously  the  steady  state  is  represented  by  the  final  charge 
on  the  condenser.  The  particular  integral,  u,  is  therefore  equal 
to  EC.  The  transient  Y  is  found  as  before  by  putting  E  =  0. 
The  solution  of  the  equation  is  of  the  form  q  =  Atat. 


138  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Substituting    E  =  0    and    q  =  Aeat   in    equation    (48)    gives 

0  =  raAeat  +  A  ^  (49) 

O 

This  holds  for  all  values  of  t. 

0  =  ra  +  ^  (50) 

Equation  (50)  is  an  equation  of  the  first  degree  and  therefore 
has  but  a  single  root. 

1 
a=     ~Cr 

The  complete  solution  of  equation  (48)  is  therefore 

q  =  Ae~^  +  EC  (51) 

The  constant  of  integration  is  found  by  putting  t  equal  to 
zero.  Let  QQ  be  the  charge  on  the  condenser  when  the  circuit 
is  closed,  i.e.,  when  t  is  zero.  Then 

A  =  Qo  -  EC  (52) 

Putting  this  value  of  the  constant  of  integreation,  A,  in  equa- 
tion (51)  gives 

q  =  EC  +  (Qo-EC)e    &  (53) 

If  the  initial  charge  is  zero,  equation  (53)  becomes 


q  = 

=  o(l-e~^)  (54) 

Where  Q  =  EC  is  the  final  charge  on  the  condenser. 
When  the  initial  charge  on  the  condenser  is  zero,  the  counter 
electromotive  force  of  the  condenser  is  zero  when  t  =  0.     The 

TjJ 

current  in  the  circuit  at  this  instant  has  its  maximum  value.  —  . 

t  r 

As  t  increases  q  also  increases,  producing  a  counter  electromotive 
force  g,  which  cuts  down  the  current  and  hence  the  rate  at  which 

the  charge  increases,  and  therefore  diminishes  the  rate  of  decrease 
of  the  current.  The  charge  approaches  the  limit  Q  =  EC,  while 
the  current  approaches  the  limit  zero. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       139 


The  rate  of  increase  of  charge  and  therefore  the  rate  of  decrease 
in  current  is  determined  by  Cr,  the  time  constant  of  the  circuit. 
Like  the  time  constant  for  a  circuit  containing  constant  resistance 
and  constant  self-inductance  in  series,  it  is  equal  to  the  time  re- 
quired for  the  charge  to  reach  0.632  of  its  final  or  maximum  value. 

Since  i  =  -¥,  the  equation  for  current  may  be  found  by  differ- 
entiating equation  (53)  with  respect  to  time. 


.  .  _  (fe±ff !.-« 


_ 

C/7* 


(55) 
(56) 


When  the  initial  charge  on  the  condenser  is  zero,  equations  (55) 
and  (56)  reduce  to 


— 6 

r 


(57) 


When  the  initial  charge  on  the  condenser  is  zero,  the  charge 
on  the  condenser  increases  logarithmically  to  the  limiting  value 
EC  (equation  (54)).  The 
current,  on  the  other  hand, 
reaches  its  maximum  value 

Tjl 

—  instantly  and  then  di- 
minishes logarithmically  to 
zero. 

The  growth  of  the  charge 
and  the  decay  of  the  cur- 
rent, when  the  initial 
charge  is  zero,  are  shown 
in  Fig.  40.  The  curves  are 
all  for  the  same  impressed 
voltage,  E,  and  the  same  resistance,  r.  The  capacitance  is  least 
for  curve  a  and  greatest  for  curve  c. 

Energy  of  the  Electrostatic  Field. — From  the  energy  equation 
(equation  (46),  page  137),  it  may  be  seen  that  energy  is  constantly 
being  dissipated  in  heat  and  also  is  being  stored  in  the  electro- 
static field  at  a  rate,  £  i,  which  rate  constantly  decreases.  The 


Time 
FIG.   40. 


140  PRINCIPLES  OF  ALTERNATING  CURRENTS 

total  energy  thus  stored  in  the  electrostatic  field  is  evidently 
given  by 

(58) 


where  Q  and  E  are  respectively  the  final  charge  and  the  final 
voltage  of  the  condenser.  The  expression  ^CE2  represents  the 
electropotential  energy  of  the  condenser  due  to  its  charge.  It 
corresponds  to  the  potential  energy  in  mechanics  of  a  stretched 
spring  or  other  elastic  body  which  is  under  stress. 

When  a  condenser  is  charged  through  a  fixed  resistance  from 
a  constant  potential,  one-half  of  the  energy  given  to  the  circuit 
is  absorbed  as  i2rt  loss  or  joule  heating  in  the  resistance.  The 
other  half  is  stored  as  electropotential  energy  in  the  condenser. 
The  efficiency  of  a  system  involving  the  charging  of  a  condenser 
from  a  constant  potential  source  through  a  fixed  resistance 
therefore  cannot  be  greater  than  fifty  per  cent. 

Putting  T,  in  the  energy  equation  (equation  (47),  page  137) 
equal  to  infinity  gives 

E\    idt  =    \    ri*dt+^  I   qidt 
Jo  Jo  C  Jo 

r^F  -JL         r°°  /F\2  —*L        1  r® 

El     -€    *dt  =  r\     (-)  e    Crdt  +  ~\    qdq 
Jo    r  Jo    \r/  CJ0  *  * 


E 


E 


_         1= 


Cr\ 
—  —  ) 

27 


2CJ  o 
E*C  =  ~  +  ~  (59) 


The  first  term  of  equation  (59)  is  the  energy  supplied  to  the 
circuit.  The  second  and  third  terms  are  respectively  the  energy 
loss  in  the  resistance  and  the  energy  stored  in  the  condenser. 
The  second  and  third  terms  are  each  equal  to  one-half  the  total 
energy  supplied  to  the  circuit. 

Case  II. — Decay  of  the  Charge  in  a  Capacitive  Circuit  when 
the  Impressed  Electromotive  Force  is  Removed  by  Short  Circuiting. — 
(Fig.  39,  page  137,  key  on  6.)  Since  electromotive  force,  e,  is 
equal  to  zero 

0-H  +      -r      +  (60) 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       141 

In  this  case  the  steady  charge  is  obviously  zero  since  the  elec- 
tromotive force  is  zero.  The  term  which  represents  the  steady 
state  in  the  equation  for  the  charge  is  therefore  zero. 

q  =  Y  +  u 

=  Y  +  0  (61) 

The  complete  solution  of  equation  (60)  consists  of  a  transient 

term  only.     The  solution  is 

_  t 
q  =  Ae~c~r 

When  t  is  equal  to  zero,  q  is  equal  to  Qo ',  the  initial  charge 
on  the  condenser,  i.e.,  the  charge  at  the  instant  it  begins  to  dis- 
charge. Therefore 

A  =Q0' 
and 

_  J. 
q  =  Qo'  e    CT  (62) 

If  the  charge  has  reached  its  final  value  EC,  where  E  is  the 
charging  potential,  before  the  electromotive  force  is  removed, 

that  is,  before  the  discharge  is  started,  equation  (62)  becomes 

__t_ 
q  =  EC  e   cr  (63) 

In  this  case  when  t  =  0,  i.e.,  at  the  instant  of  short  circuiting 
or  removing  the  electromotive  force,  the  charge  q  is  equal  to 
Qo'  =  EC.  The  charge  decreases  logarithmically  to  zero. 

Since  i  =  -j.,  the   equation  for  current  may  be  obtained  by 

differentiating  the  equation  for  charge,  i.e.,  equation  (62),  with 
respect  to  time. 

i  =  ~^r€~^r  (64) 

When  Qo'  =  EC,  i.e.,  when  the  steady  state  has  been  reached 
before  the  condenser  begins  to  discharge,  the  expression  for  the 
current  becomes 

E  __L 

I    = €     Cr 

r 

=  -/o'e~^  (65) 

In  this  case  the  current  i  is  equal  to  its  Ohm's  law  value, 

/o'  =  — ,  when  t  is  zero. 


142 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


At  the  instant  of  removing  the  electromotive  force,  i.e.,  when 
t  is  zero,  the  charge  on  the  condenser  is  a  maximum,  being  equal 
to  Q  =  EC,  where  E  is  the  voltage  of  the  condenser  at  the  instant 
discharge  is  started.  As  t  increases,  the  charge  diminishes 
logarithmically  to  zero.  The  current  has  its  maximum  value, 

— ,  for  t  =  0  and  also  diminishes  logarithmically,  approaching 

zero  as  a  limit  with  the  charge.  Observe,  however,  that  the 

current  is  negative,  which 
means  that  the  current  is 
flowing  out  of  the  condenser. 
During  the  discharge  the 
energy  of  the  electrostatic 
field  is  being  gradually  dis- 
sipated in  heat  in  the  re- 
sistance of  the  circuit. 

The  decay  of  charge  and 
current  are  shown  graphically 
in  Fig.  41.  The  capacitance 
is  least  for  curve  a  and  greatest 
for  curve  c. 

Case  III.— Simple  Harmonic 
Electromotive  Force  Impressed 
on  a  Circuit  Containing  Con- 
stant Resistance  and  Constant 
Capacitance  in  Series. — In  this 
case  the  electromotive  force 
is  e  —  Em  sin  (coZ  +  a)  and 
FIG.  41.  equation  (45),  page  136 

becomes 

E»      •     /  j  dq    .    q  ,     . 

&m  sin  (cot  -|-  a)  =  r  ,    -f-  ^  (bo) 

This  is  a  linear  differential  equation  of  the  first  order  with 
constant  coefficients.  The  first  term  is  a  function  of  t.  The 
solution  is  of  the  form 

q  =  Y  +  u 

The  complementary  function,  Y,  is  found  in  the  same  manner 
as  is  Case  I,  by  putting  the  left-hand  member  of  the  equation 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       143 

equal  to  zero.  This  gives  the  transient  component  of  the  charge. 
From  equation  (51)  this  is 

Y  =  A  6~c~'  (67) 

The  particular  integral,  i.e.,  the  term  representing  the  steady 
state,  may  be  found  by  a  method  similar  to  that  used  in  Case  III, 
page  128,  for  an  inductive  circuit  with  a  sinusoidal  electromotive 
force  impressed. 

Under  steady  conditions,  the  voltage  drop  across  the  circuit 

da 
will  be  made  up  of  two  parts:  one,  ri  =  r-r.,  to  supply  the 

resistance   drop;  the   other,   ^,  caused  by  the   charge  on  the 

condenser. 

When  the  charge  varies  sinusoidally  with  time,  the  resistance 
drop  will  also  be  sinusoidal  in  form,  provided  the  resistance  is 
constant,  for  then  the  drop  will  be  equal  to  a  constant  multiplied 
by  the  derivative  of  a  sine  function.  The  derivative  of  a  sine 
function  is  a  cosine  function,  which  is  equivalent  to  a  sine  function 
advanced  90  degrees  in  phase.  If  C  is  constant,  the  voltage  drop, 

p  across  the  condenser  terminals  will  obviously  be  of  the  same 

wave  form  as  the  charge.  Since  the  sum  of  any  number  of 
sinusoidal  waves  of  like  frequency  is  a  sinusoidal  wave  of  the 
same  frequency,  it  follows  that  if  r  and  C  are  both  constant  and 
the  charge  varies  sinusoidally  with  time,  the  voltage  across  the 
circuit  required  to  produce  the  charge  must  also  be  sinusoidal  in 
wave  form,  since  it  is  the  sum  of  two  sinusoidal  drops  related 
to  the  charge.  Conversely,  if  the  voltage  impressed  on  the 
circuit  is  sinusoidal,  the  sum  of  the  reactions  due  to  the  charge 
must  be  sinusoidal.  The  only  way  the  sum  of  these  reactions  can 
be  sinusoidal,  when  both  r  and  C  are  constant,  is  for  the  charge  to 
be  sinusoidal. 

Let  the  charge  be 

q  =  Qm  sin  (orf  +  a  +  0')  (68) 

where  Qm  is  the  maximum  value  of  the  charge  after  steady  condi- 
tions have  been  attained  and  0'  is  the  phase  angle  of  the  charge 
with  respect  to  the  impressed  electromotive  force  Em. 


144 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Substituting  the  value  of  the  charge  given  by  equation  (68) 
in  equation  (66)  gives 

d  ,v, 

Em  sin  (co£  +  a)  =  r  -r  \Qm  sin  (co£  +  a  +  6  )  \ 

+  ^Qm  sin  (ol  +  a  +  0') 
=  rcoQm  cos  (co£  +  or  +  0') 

+  ~  Qm  sin  («<  +  a  +  0')  (69) 

Since  a  cosine  function  leads  a  sine  function  by  90  degrees, 
equation  (69)  may  be  written 

Em  sin  (coZ  +  a)  =  ruQm  sin  (cot  +  a  +  0'  +  90°) 
1 


sn 


+  a  + 


(70) 


The  two  components  of  the  second  member  of  equation  (70) 
are  quadrature  components  of  the  voltage  drop,  Em  sin  (to£  +  a), 

across  the  circuit.     The  maximum 
Em(cosd'+jsmd')   values    of    these    components    are 

ruQm   and  -£• 

The  vectors  corresponding  to  the 
three  terms  of  equation  (70)  are 
plotted  in  Fig.  42  for  the  instant  of 

Qm  time  t  = —  •     At  this  instant 

C*  CO 

FIG.  42.  the  vector  representing  the  charge 

lies  along  the  axis  of  reals. 
Referring  to  Fig.  42,  it  is  obvious  that 


Em  = 


V 


cox  r2  + 


=  tan"1  —    =  tan  -1  o>rC 


(71) 

(72) 

(73) 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       145 

The  voltage  Em  leads  the  charge  Qm  by  an  angle  6'  or  the 
charge  lags  the  voltage  by  the  same  angle.  Since  0'  in  equation 
(70)  is  the  phase  angle  of  the  charge  with  respect  to  the  voltage, 
it  should  be  considered  negative,  as  the  charge  lags  the  voltage. 

The  particular  integral  or  steady  term  for  the  charge  in  a 
circuit  containing  constant  resistance  and  constant  capacitance 
in  series  is  consequently 

q  =  —    Em-    =  sin  M  +«  -  tan"1  rcoC)  (74) 


The  complete  solution  of  equation  (66)  is  therefore 

-—  F 

q  =  Ae   Cr  +  -  =  sin  M  +  «  -  tan-1  ro>C)     (75) 


If  the  initial  charge  is  zero,  i.e.,  the  condenser  is  uncharged 
when  the  circuit  is  closed,  q  will  be  zero  when  t  is  zero.  .  Putting 
both  t  and  q  zero  in  equation  (75)  gives 

~==  =  sin  (a  -  tan-1  rcoC)  (76) 


Therefore 


—  E  -- 

m         sin  (a  -  tan-1  ro>C)e   Cr 


sin  (ut  +  a  —  tan"1  r<oC)  (77) 


C02C2 

Since  i  =  -r,  the  equation  for  the  current,  when  a  sinusoidal 

electromotive  force  is  impressed  on  a  circuit  containing  constant 
resistance  and  constant  capacitance  in  series,  may  be  found  by 
differentiating  equation  (77)  with  respect  to  t. 

v  « 

sin  (a  —  tan"1  ro>C)€    Cr 


cos  («*  +  a  -  tan-1  rcoC)      (78) 


10 


146  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Replacing  6'  by  90°  —  0  (see  Fig.  42)  and  also  remembering 
that  cos  (|8  —  90°)  =  sin  0  and  sin  (0  —  90°)  =   —cos  /3,  gives 

-%_       ,  cos  (a  +  tan'1     l  \   ~^r 


1    o>2C2 

Em 

(79) 


where  tan"1  —  ~  =  0  is  the  angle  of  lead  of  the  current  with  respect 

to  the  impressed  voltage  after  steady  conditions  have  been 
established. 

It  should  be  observed  that  0  is  positive  and  therefore,  after 
steady  conditions  have  been  reached,  the  current  leads  the 
impressed  electromotive  force  in  a  circuit  containing  constant 
resistance  and  constant  capacitance  in  series  by  an  angle  whose 
tangent  depends  upon  the  resistance,  capacitance  and  frequency 
of  the  circuit.  For  a  circuit  containing  constant  resistance  and 
constant  self-inductance  in  series  the  current  lags  the  impressed 
electromotive  force  by  an  angle  whose  tangent  depends  upon  the 
resistance,  self-inductance  and  frequency  of  the  circuit.  (See 
page  131.) 

The  limiting  value  a  current  can  attain  when  a  sinusoidal 
electromotive  force  is  impressed  on  a  circuit  containing  constant 
resistance  and  constant  self-inductance  in  series  is  twice  its 
steady  state  maximum  value.  (See  page  132.)  No  such  limita- 
tion exists  when  a  sinusoidal  electromotive  force  is  impressed  on  a 
circuit  containing  constant  resistance  and  constant  capacitance 
in  series.  For  fixed  r  and  C,  the  transient  will  be  a  maximum 
when  the  circuit  is  closed  at  a  point  on  the  electromotive  force 

wave  which  will  make  (a  +  tan"1  —  ^j  =  0.    (See  equation  (79).) 

Under  this  condition  the  steady  term  is  zero  when  t  is  zero,  that  is 
at  the  instant  the  circuit  is  closed.  The  transient  has  its  greatest 
value  when  t  =  0,  that  is,  at  the  instant  the  circuit  is  closed  and 
will  then  decrease  logarithmically  to  zero  at  a  rate  which  will 
depend  upon  the  time  constant,  Cr. 

The  ratio  of  the  maximum  value  of  the  transient  term  to  the 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       147 

maximum  value  of  the  steady  term  in  the  current  equation  (equa- 
tion (79))  increases  as  the  ratio  of  ^  =  x  to  r  increases  and  ap- 
proaches infinity  as  a  limit.  The  maximum  value  of  the  transient 

J? 
term  approaches  the  Ohm's  law  value  of  the  current,  i.e.,  — -> 

as  the  ratio  of  —^  to  r  increases,  since  the  coefficient  of  the 

T? 
transient  term  becomes  approximately  equal  to  — -  when  r  is 

small  compared  with  — ^* 

The  ratio  of  r  to  -7,  at  sixty  cycles  for  a  well  made  commercial 
static  condenser  may  be  as  low  as  0.01  or  0.02.     If  a  condenser 

having  a  ratio  of  r  to  -7,  equal  to  0.01  were  connected  to  a  con- 
coC/ 

stant  potential,  60-cycle  circuit  at  the  point  on  the  electromotive 
force  wave  which  would  make  the  transient  a  maximum,  the 
current  would  immediately  rise  to  approximately  100  times  the 
value  it  would  have  after  steady  conditions  were  established. 
For  circuits  having  constants  likely  to  be  met  in  practice, 
the  transient,  although  it  may  be  initially  very  large,  practically 
disappears  in  a  very  small  fraction  of  a  cycle.  For  example :  in 
the  case  just  mentioned,  o>  would  be  377.  Let  r  be  0.1  ohm. 
Then 

--  =  0.01 


For  the  transient  to  fall  to  0.1  its  initial  value 

e"^  =  0.1 


The  duration  of  the  transient  is  too  short  to  be  of  importance. 


148  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  maximum  value  of  the  steady  component  of  the  current 
is  found  by  dividing  the  maximum  value  of  the  voltage  by 

Jr2  _j  —  *  .  it  leads  the  maximum  value  of  the  voltage  by  an 
angle  whose  tangent  is  equal  to  the  ratio  of  —  ^  to  r,  i.e.,  by  an 

angle  whose  tangent  is  —  ^-     Since  the  effective  value  of  a 

sinusoidal  current  is  equal  to  its  maximum  value  divided  by  the 
square  root  of  two,  it  is  evident  that  the  effective  value  of  the 
current  is  given  by  the  effective  value  of  the  voltage  divided  by 

r2  _|  —  _.     The  quantity  <*/r2  -\  —  ^  plays  the  same  part  in  a 


capacitive   circuit  that   the   quantity   \/r2  +  co2L2  does  in   an 
inductive  circuit. 


Impedance  and  Reactance.  —  The  quantity 


'      C02C2 

is  called  the  impedance  and  is  measured  in  ohms.  It  is  a  constant 
only  when  resistance,  capacitance  and  frequency  are  constant. 

The  expression ^  is  called  the  capacitive  reactance  and  is 

coO 

also  measured  in  ohms.  Reactance  is  denoted  by  the  letter  x. 
Capacitive  reactance  is  constant  only  when  frequency  and 
capacitance  are  constant.  The  capacitance  of  a  circuit  under 
ordinary  conditions  is  constant. 

It  should  be  noted  that  inductive  reactance 

XL  =  coL  =  2irfL 
is    directly    proportional    to  frequency.     Capacitive  reactance, 

-—  — 

on  the  other  hand,  is  inversely  proportional  to  frequency.  In- 
ductive reactance  is  positive.  Capacitive  reactance  is  negative. 
The  significance  of  the  negative  sign  with  capacitive  reactance 
will  be  understood  from  the  "Vector  Method"  which  follows. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        149 


Vector  Method  of  Determining  the  Steady  Component  of  the 
Current  in  a  Circuit  Having  Constant  Resistance  and  Constant 
Capacitance  in  Series. — The  current  vector,  7  amperes  effective, 
will  be  taken  along  the  axis  of  reals.  The  effective  values  of  the 
active  and  reactive  components  of  the  impressed  electromotive 

force  are  given  by  rl  and  —j~r  ^  ~  ~~JXI  respectively.     The 

reactive  component  must  be  negative  since  the  current  in  a 
capacitive    circuit    leads    the    impressed    electromotive    force. 


E  =  rl  -jxl  =  I(r  -jx)  =  Iz 

The  vectors  corresponding  to  equa- 
tion (81)  are  plotted  in  Fig.  43. 

The  waves  corresponding  to  the 
vectors  in  Fig.  43  are  shown  in  Fig. 
44.  Time  is  taken  zero  when  the 

current  is  zero. 

_j  — 

In  complex,  therefore,  the  impedance      w 
of  a  capacitive  circuit  is  given  by 

z  =  r  —  jx 


(81) 


E  (cos  8-j  sin  0) 
FIG.  43. 


(82) 


FIG.  44. 


The  value  of  z  in  ohms  is,  of  course, 


Calling  the 


component  of  E  along  the  axis  of  reals,  i.e.,  the  real  or  active 
component,  Ea,  and  the  component  along  the  axis  of  imaginaries, 
i.e.,  the  imaginary  or  reactive  component,  Er,  gives 


150  PRINCIPLES  OF  ALTERNATING  CURRENTS 

E  =  Ea  -  jEr 

E  (Volts)    =    V#a'  +  Er2 


I  =  Ea  ~  jEr  (83) 

r  -  jx 

Rationalizing  equation  (83),  to  get  rid  of  j  in  the  denominator, 
by  multiplying  the  numerator  and  the  denominator  by  the 
denominator  with  the  sign  of  the  term  involving  j  reversed, 
gives 

7  =  Ea  -  jEr       r+jx 
r  —  jx          r  +  jx 
_  Ear  +  Erx  __    Err  -  Eax  _ 

2  2  3         2  2 


where 


,     _  Ear  +  Erx  Err  -  Eax 

--  ~~ 


are  respectively  the  active  and  reactive  components  of  the 
current.  But  7  is  along  the  axis  of  reals.  Therefore  Ir  =  0. 
Hence,  Err  =  EaX  and 

|r  =  x  =  tan  e  (85) 

-H/a          T 

where  6  is  the  angle  of  lead  of  the  current  with  respect  to  the 
voltage  impressed  across  the  circuit.  The  power  is  Eala  =  El 
cos  6. 

Polar  Expression  for  the  Impedance  of  a  Circuit  Containing 
Constant  Resistance  and  Constant  Capacitance  in  Series.  —  The 
polar  expression  for  the  impedance  of  a  circuit  containing  constant 
resistance  and  constant  capacitance  in  series  is  (see  page  136) 

z  =  z\-B  =  zH  (86) 

where  the  angle  6  is  determined  by  the  relation 

6  =  tan-1  * 

For  capacitance  x  is  negative.  Therefore  0,  in  the  polar 
expression  for  the  impedance  of  a  circuit  containing  resistance 
and  capacitance  in  series,  is  negative. 

Circuit  Containing  Constant  Resistance,  Constant  Self- 
inductance  and  Constant  Capacitance  in  Series.  —  The  conditions 
in  a  circuit  containing  constant  resistance,  constant  self-indue- 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       151 

tance  and  constant  capacitance  in  series  are  completely  deter- 
mined by  the  equation 

e  =  ri  +  L—   +  -  (87) 

where  e  is  the  impressed  electromotive  force  drop  and  ri,  L^r. 

and  ^  are,  respectively,  the  components  of  the  impressed  electro- 
motive force  drop  to  supply  the  ohmic  drop,  the  drop  due  to 
self-induction  and  the  drop  due  to  the  condenser.  If  the  electro- 
motive force,  6,  is  in  volts,  r  must  be  in  ohms,  L  in  henrys  and 
C  in  farads.  The  current  and  charge  will  then  be  in  amperes 
and  coulombs  respectively. 

The  energy  relation  corresponding  to  equation  (87)  is 


fTeidt  =    (Tri*dt  +   f 


(88) 


The  first  term  of  the  second  member  of  equation  (88)  is  the 
energy  dissipated  in  heat  in  the  resistance  of  the  circuit.  The 
second  term  is  the  energy  stored  in  the  magnetic  field  of  the 
inductance  and  the  third  term  is  the  energy  stored  in  the  electro- 
static field  of  the  condenser. 

Solution  of  the  Differential 
Equation  for  a  Circuit  Containing 
Constant  Resistance,  Constant 
Self-inductance  and  Constant 
Capacitance  in  Series. — The  solu- 
tion of  equation  (87)  for  current  in 
terms  of  e,  r,  L,  C  and  t  will  evidently 
depend  for  its  form  upon  e. 

Case  L — Charge  of  a  Condenser  through  a  Circuit  Containing 
Constant  Resistance  and  Constant  Self-inductance  in  Series  on 
which  is  Impressed  a  Constant  Electromotive  Force. — (Fig.  45, 
key  on  a.) 

E  =  ri  +  Ld4.  +  I-  (89) 


FIG.  45. 


(90) 


152  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Equation  (90)  is  a  linear  differential  equation  of  the  second 
order  with  constant  coefficients.  Its  complete  solution  for 
charge,  q,  is  of  the  form 

q  =  Y  +  u  (91) 

where,  as  in  all  previous  cases,  Y  and  u  are,  respectively,  the 
complementary  function  and  particular  integral  and  represent 
the  transient  and  steady  states.  As  in  Case  I,  for  a  circuit 
having  constant  resistance  and  constant  capacitance  in  series, 
the  current  will  be  zero  when  the  steady  state  is  reached.  Under 
this  condition  the  charge  on  the  condenser  will  obviously  be 
constant  and  equal  to  CE.  The  term  u  in  equation  (91),  repre- 
senting the  steady  state,  is  therefore  CE.  The  term  Y,  repre- 
senting the  complementary  function  or  transient,  is  found  by 
putting  E  =  0  and  q  =  eat.  (See  Murray,  Differential  Equa- 
tions, page  63,  second  edition.) 
Making  these  substitutions  gives 


0  =     ra+La2  +     rea<  (92) 

This  holds  for  all  values  of  t. 

0  =  ra  +  La2  +  g  (93) 

Equation  (93)  is  of  the  second  degree  and  therefore  has  two 
roots.     These  are 


Bi  _  -re  +  (94) 


and 

*  7~2™  (95) 


The  complete  solution  of  equation  (90)  is  therefore 

q  =  A^  +A2ea^  +EC  (96) 

where  A\  and  A2  are  arbitrary  constants  of  integration. 

0.         .        dq 

Since  i  =  ~ 

at 

i  =  Aiaieait  +Ata2ea2t  (97) 


RESISTANCE,  .INDUCTANCE  AND  CAPACITANCE       153 

The  form  taken  by  the  solutions  of  equations  (96)  and  (97) 
depends  on  the  relation  of  r2C2  to  4Z/C.  If  r2C  is  greater  than  4L, 
the  roots  a\  and  a2  are  both  real.  If  r2C  is  less  than  4L,  both  roots 
are  imaginary.  If  r2C  is  equal  to  4L,  the  roots  are  equal. 


Cose  /-a. —  r2C  >  4L  or  r  >  2-/g. — In  this  case  it  is  evident 

that  the  two  roots  ai  and  a*  are  essentially  negative.  If  t  =  0, 
q  =  0  and  i  =  0.  Putting  these  values  in  equations  (96)  and 
(97)  and  solving  for  the  constants  AI  and  A2  gives 


01  -  a2  2       2C2  -  4LC 


Hence,  from  equations  (96)  and  97) 

(rC  + Vr2C2-4LC    -("        2LC 

q  =  tiL  —  .frC        .  —  € 

n  -  '  °C2  -  4LC 


-4LC 

7»l 

(99) 


-4LC€  2LC  y-i    (100) 


-  4LC 


I    = 


i?r          \     frC  ~  v>-tga-4z>cNj 
^ L~( 2LC— 

frC  +  Vr*C*  -  4LCN t 

(101) 


V2C2  -  4LC , 

~\  2LC 


—  € 


The  charge  on  the  condenser,  starting  at  the  value  zero, 
approaches  the  value  Q  =  EC  as  its  limit.  The  current,  on  the 
other  hand,  starts  at  the  value  zero,  rises  to  a  maximum  and  then 
decreases,  approaching  zero  as  its  limit. 

do 

Since  i  =  TT,  it  is  evident  that  the  slope  of  the  curve  repre- 
senting the  charge  is  proportional  to  the  current  at  any  instant. 
This  curve  must,  therefore,  have  a  point  of  inflection  at  the  time 


154 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


when  the  current  has  its  maximum  value.  In  Fig.  46  are  given 
the  curves  of  charge  and  current  for  the  following  values  of  the 
constants. 


E  =  2000  volts. 
r  =  100  ohms. 


L  =  0.0016  henry. 
C  =  1  microfarad. 


0.0001     Second      0.0002 

FIG.  46. 


0.002 


Case  I-b. —  r2C  <  4L  or  r  <  2x/^. —  In  this  case  it  is  evident  that 

the  roots  ai  and  a2,  equations  (94)  and  (95),  page  152,  are  imagi- 
nary, since  the  radical  in  the  expressions  for  both  a\  and  a2  is 
negative  and  may  be  written  A/—  1  X  \/4LC  —  r2C2. 

Using  the  operator  j  to  indicate  the  imaginary,  -\/  —  1,  the 
expressions  for  a\  and  a2  become 


and 


.  V4LC  -  r2C'< 
1          2LC 


r        .  V4LC  -  r*C' 


~2L~J 


2LC 


=  a+jb 


=  a  —  jb 


(102) 


(103) 


r                                  -  r2C2      _ 
Here  a  =    —        anc*  o  =  -    — ^ •     Both  a  and  6  are 


real.     Equation  (96),  page  152,  now  becomes 

q  =  A,e(n+ib)t  +  A*(n-jb]t  +EC 


(104) 
(105) 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       155 

From  the  relations  (cos  6  +  j  sin  0)  =  <?e  and  (cos  6  —  j  sin  6) 
=  e  je  (see  pages  15  and  16)   may  now  be  written 
q  =  c°*{Ai(cos  bt  +j  sin  bt)  +  A2(cos  bt  -j  sin  bt)}  +  EC 
=  (Ai  +  AJe"  cos  bt  +j(Ai  -  A2)eat  sin  bt  +  EC 
=  Ae°'  cos  bt  +  B^  sin  bt  +  EC  (106) 

in  which  A  =  (Ai  +  A2)  and  B  =  j(Ai  —  A2) 

From  equation  (106),  since  *  =  j? 


i  =  (Aa  +  5&)e°<  cos  fa  +  (Ba  -  A&)ea'  sin  &*  (107) 

If  t  =  0,  5  =  0  and  i  =  0.     Therefore 
A  =  -EC 


*>          V4LC  -  r2C2 
and 


EC  +  e"  Al  +      sin 


in  (bt  - 

+ta"-l(-a)! 

=  EC  -  ECf'Jl  +  -  sin    bt  +  tan"1  ( -   j) 

2\/LC         -£-         <V±LC  -r2C2 

=  EC  —  EC     /  e    tt  sm  — i 

V4LCT-  r2C2  2LC 

+  tan~l ^p  (108) 

From  equation  (107)  the  expression  for  the  current  becomes 


oET/nr  _  rt 

i   =  ^     g     2L  gin  ^_±1^ L_±_  i  (109) 

A/4LC  -  r2C2  2LC 

It  is  evident  that  the  charge  and  the  current  have  the  same 
initial  and  final  values  as  in  Case  7-a,  just  discussed.  There 
is,  however,  an  oscillation  about  these  values,  the  amplitude  of  the 
oscillations  decreasing  logarithmically.  The  period  of  these 

\/4LC  —  r2C2 
oscillations  is  determined  by  the  fact  that  when  -         rri      —  t 


156  PRINCIPLES  OF  ALTERNATING  CURRENTS 

increases  by  2ir,  the  charge  and  current  pass  through  one  com- 
plete cycle  of  values.     Hence  the  increase  in  time,  t',  is  given  by 


t' 


2LC 
Therefore 

T  —  t'  - 

~^C  (H°) 

4L 

r2C 

If  r2C  is  very  small  compared  to  4L,  so  that  -77-  may  be  neg- 
lected, T,  the  time  of  a  complete  period,  is 

T  =  27r\/LC  (111) 

r2C 
As  "Ty-  approaches  unity  the  periodic  time,    T,   approaches 

infinity.     (See  equation  110.) 

Fig.  47  shows  equations  (108)  and  (109)  plotted  for  the  follow- 
ing constants : 

E  =  2000  volts.  L  =  0.0125  henry. 

r  =  100  ohms.  C  =  1  microfarad. 


FIG.  47. 


Case  I-c. —  r2C  =  4L  or  r.  =  2^/— . — In  this  case  the  roots 

and  a2,  equations  (94)  and  (95),  page  152,  are  equal,  i.e.,  ai  =  a2 
a  =  —  2j.     Equations  (96)  and  (97)  now  become 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       157 

q  =  A^at  +  A2teat  +  EC*  (112) 

and 

i  =  Aiaf*  +  Azat^  +  A2e°'  (113) 

If  t  =  0,  q  =  0  and  i  =  0.     Putting  these  values  in  equations 
(112)  and  (113)  gives 

Ax  =  -EC 


=  aEC  =  -^ 


and 


q  =  EC  - 


i* 

~2L 


(114) 
(115) 


It  is  important  to  note  that  in  this  ©d,se,  as  in  Case  /-a,  both 
charge  and  current  are  non-oscillatoryX 

In  Fig.  48  are  shown  graphically^the  curves  for  charge  and 
current  for  the  following  constants: 

E  =  2000  volts.  L  =  2.5  henrys. 

r  =  100  ohms.  C  =  1000  microfarads. 


Second 


0.06 


0.10 


0.15 


0.20 


FIG.  48. 


Case  II.  —  Discharge  of  a  Condenser  through  a  Circuit  Contain- 
ing Constant  Resistance  and  Constant  Self-inductance  in  Series.  — 
(Fig.  45,  page  151,  key  on  6.)  The  differential  equation  for 
this  condition  is 


a -**:££  +  * 


'<fc   '    C 
*  See  Murray,  Differential  Equations,  page  65,  Second  Edition. 


(116) 


158  PRINCIPLES  OF  ALTERNATING  CURRENTS 

or 


In  this  case  the  final  or  steady  state  will  be  zero  charge, 
since  the  applied  voltage  is  zero.  The  equation  for  charge  will 
contain  only  a  transient  term.  The  steady  term,  u,  in  equation 
(61),  page  141,  becomes  zero  since  the  final  charge  is  zero. 

q  =  A,eait  +  A2ea2i  (118) 


«.  .       do 

Since  i  =  -~ 

at 


= 


lt  +  A2a2ea2t  •   (119) 


The  roots  a\  and  a2  will  have  the  same  values  as  in  Case  I, 
equations  (94)  and  (95),  page  152.  The  method  of  solving 
equations  (118)  and  (119)  is  the  same  as  was  used  for  finding  the 
transient  terms  of  equations  (96)  and  (97)  under  Case  /,  page  152. 
As  before,  there  are  three  cases  to  consider,  according  as  r2C  is 
greater  than  4Z/,  less  than  4L  or  equal  to  4L. 

Case  Il-a. —   r2C  >  4L  or  r  >  2*/^. — In  this  case  the  roots 

ai  and  az  are  essentially  negative  and  have  the  same  values  as  in 
Case  I -a,  page  153 


-(-:-*nz-  -)' 

-  4LC 

rC  -  V^^^LC  ^  ~(~    -^~     JC>         (12o) 

and 


-EC 

e 


-  4LC 


_  (rC  -   Vr*C2  -  4LC\ 
V  2LC  ) 


-4LC)t 

\  -   AJ\^f 

—  e 


2LC  [  (121) 


In  Fig.  49  are  shown  graphically  the  curves  of  charge  and 
current  during  the  discharge  of  a  condenser  through  a  circuit 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       159 


having  constant  resistance  and  constant  self-inductance  in  series 
for  the  following  constants: 

E  =  2000  volts.  L  =  0.0016  henry, 

r  =  100  ohms.  C  =  1  microfarad. 


FIG.  49. 


~ — In  this  case  as  in  Case  7-6, 


CaseII-b.—r2C  <  4Lorr  < 

page   154,  the  roots  ai  and  a%  are  imaginary  and  have  the 
same  values  as  in  that  case.     Equations  (118)  and  (119)  become 

.    f\/4LC- 


q  =  EC 


V4LC  - 


2L  sin 


2LC 


+  tan- 


rC 


-  r2C2 1 


-  2EC          -  Jl    .    V4LC  -  r2C2  . 

I  =  — e        2L  SID  —  -    t 

A/4LC  -  r2C2  2LC 


(122) 
(123) 


The  charge  and  current  have  the  same  initial  and  the  same 
final  values  as  in  Case  Il-a.  They  oscillate  about  these  final 
values  with  an  amplitude  which  decreases  logarithmically. 

The  time  of  one  complete  vibration  for  charge  or  current  is 
the  same  as  in  Case  I-b  and  is  found  in  the  same  way. 


T  = 


2-rVLC 


4L 


160  PRINCIPLES  OF  ALTERNATING  CURRENTS 

If  r2C  is  negligibly  small  as  compared  to  4L, 


T  = 

If  r2(7  =  4L,  T,  the  time  of  a  complete  vibration,  is  infinite- 
The  oscillatory  discharge  of  a  condenser  is  shown  graphically 
in  Fig.  50  for  the  following  constants: 

E  =  2000  volts.  L  =  0.0125  henry. 

r  =  100  ohms.  C  =  1  microfarad. 


J).OQ2 


0.001 


FIG.  50. 


Case  II-c. —  r2C  =  4L  or  r  =  2-J  -^. — Here  the  roots  a\  and 

a2  are  equal  as  in  Case  I-c,  page  156,  and  have  the  same  values 
as  in  that  case.  The  method  of  solving  the  differential  equation 
is  the  same  as  in  Case  I-c,  except  that  the  term  EC,  which 
represents  the  steady  condition,  is  zero.  The  solution  involves 
nothing  but  the  transient  term  since  the  final  charge  must  be 
zero.  From  equations  (112)  and  (113),  page  157, 


i  =  Aiaeat  +  A^ateat  +  A2eat 
The  solution  of  equations  (125)  and  (126)  gives 


q  = 


i  = 


2L 


-i1* 


rt 
2L 


(125) 
(126) 


(127) 
(128) 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        161 


Where  EC  =Q  and  E  are,  respectively  the  initial  charge  and 
initial  voltage  of  the  condenser.  Equations  (127)  and  (128)  are 
shown  graphically  in  Fig.  51  for  the  following  constants: 


E  =  2000  volts. 
r  =  100  ohms. 


L  =  2.5  henrys. 

C  =  1000  microfarads. 


0.10 


Second 


0.20 


FIG.  51. 


Case  III. — A  Simple  Harmonic  Electromotive  Force  Impressed 
on  a  Circuit  Containing  Constant  Resistance,  Constant  Self-in- 
ductance and  Constant  Capacitance  in  Series. — In  this  case  the 
impressed  electromotive  force  is  of  the  form 


e  =  Em  sin  (ut  -\-  a)  and 
Em  sin  (ut  +  a)  =  ri  +  l      + 


or 


(129) 


(130) 


The  complete  solution  of  this  linear  differential  equation  of  the 
second  order,  the  first  term  of  which  is  a  function  of  t,  is  the  sum 
of  the  transient  and  the  steady  value  of  the  charge.  As  in  all 
preceding  cases  the  solution  is  of  the  form 

9=  Y  +  u 

where  Y  is  the  transient  term,  i.e.,  the  complementary  function 

of  the  differential  equation,  and  u  is  the  particular  integral  and 
11 


162  PRINCIPLES  OF  ALTERNATING  CURRENTS 

represents  the  steady  state.  The  transient  term  or  comple- 
mentary function  is  found,  as  in  all  other  cases,  by  putting  the 
first  term,  Em  sin  (ut  +  a),  equal  to  zero.  The  value  of  the 
complementary  function  evidently  depends  on  the  relation  of  r2C 
to  4L.  Its  value  is  of  the  same  form  as  in  Cases  Il-a,  b  and  c. 
It  is 

q  =  A^  +  A2ea2<  (131) 

i  =  A^e111  +  A2a2ea2t  (132) 

It  is  oscillatory  when  r2C  is  less  than  4L  or  r  is  less  than  2*  /  ^> 

\  C 

the  frequency  being  given  by  (equation  (124),  page  159). 


f  =       =  j4f  (133) 

~  T  ~      27TA/IC 

which  is  determined  by  the  constants  of  the  circuit.  The  fre- 
quency is  entirely  independent  of  the  frequency  of  the  impressed 
voltage.  The  constants  for  the  transient  terms  for  charge  and 
current  when  the  impressed  voltage  is  sinusoidal  are  determined 
from  the  charge  and  the  current  when  t  =  0.  When  the  tran- 
sient terms  are  oscillatory,  the  constants  are  determined  from 
equations  (106)  and  (107),  page  155. 

The  transient  is  of  importance  in  all  switching  operations  on 
circuits  containing  resistance,  self-inductance  and  capacitance 
in  series,  since  dangerous  oscillations  may  be  produced  if  the 

resistance   of  the   circuit  is  low    compared   with   2*/—     It  is 

important  in  all  radio  work  involving  tuning  of  series  circuits. 
Most  radio  circuits  have  mutual-inductance  as  well  as  self-in- 
ductance, and  for  such  circuits  the  conditions  are  not  so  simple 
as  those  just  discussed. 

The  transients  in  transmission  lines  may  be  of  great  importance 
during  switching  operations,  short-circuits,  etc.,  but  the  equations 
just  developed  do  not  apply  to  a  transmission  line,  since  a  trans- 
mission line  is  not  a  simple  series  circuit.  It  is  a  circuit  contain- 
ing series  resistance  and  series  inductance,  but  it  has  parallel 
capacitance  and  leakance  between  the  conductors  and  between  the 
conductors  and  the  earth. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE        163 

The  particular  integral,  i.e.,  the  term  representing  the  steady 
state  for  equation  (130),  page  161,  may  be  found  most  easily  by 
a  method  similar  to  that  used  in  Case  III  for  a  circuit  contain- 
ing constant  resistance  and  constant  self-inductance  in  series 
and  also  in  Case  III  for  a  circuit  containing  constant  resistance 
and  constant  capacitance  in  series. 

Under  steady  conditions,  the  voltage  drop  across  the  circuit 

da 
will  be  made  up  of  three  parts,  namely  :  ri  =  r  -^>  the  voltage 

_  di        T  dzq 
drop  across  the  resistance,  L  -r  =  L~jT2'  the  voltage  drop  across 

the  self-inductance,  and    ,  =       I  i  dt,  the  voltage  drop  across  the 


=  ^  I  i 


capacitance.  If  r,  L  and  C  are  constant,  the  reasoning  that  was 
used  in  Case  III  for  a  circuit  containing  constant  resistance  and 
constant  self  -inductance  in  series  and  also  in  Case  III  for  a  circuit 
containing  constant  resistance  and  constant  capacitance  in  series 
will  show  that  both  the  charge  and  current  must  vary  sinusoi  dally 
with  time  when  the  impressed  electromotive  force  is  sinusoidal. 
Assume  that  the  charge  is  given  by 

q  =  Qm  sin  («*  +  «  +  0')  (134) 

where  tf  is  the  phase  angle  of  the  charge  with  respect  to  the 
impressed  electromotive  force.     Then 

*-  *><-  +  «>  -rg  +  Lg  +  g 

=  wrQm  cos  (ut  +  a  +  0') 

-  o>2LQm  sin  (ut  +  a  +  0') 

+  ^Qm  sin  M  +  «  +  0')      (135) 

Since  the  cosine  of  an  angle  is  equal  to  the  sine  of  ninety 
degrees  plus  the  angle,  equation  (135)  may  be  written  in  the 
following  form  : 

Em  sin  (««  +  a)  =  corQTO  sin  (coZ  +  a  +  tf  +  90°) 
-u*LQm  sin  (arf+  a  +  0') 

+     Qm  sin  («;*  +  «  +  00  (136) 


164 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


The  vectors  corresponding  to  the  terms  of  equation  (136)  are 

OL  +  Q 
plotted  in  Fig.  52  for  the  instant  of  time  t  =  — 


CO 


This 


instant  of  time  was  chosen  because  it  puts  the  vector  representing 
the  charge  along  the  axis  of  reference. 


-o>2LQ 


FIG.  52. 

From  Fig.  52  it  is  obvious  that 


Qm 


Em  = 


and 


Qm    = 


Em 


(137) 


(138) 


tan  tf 


~Qm 


(139) 


The  charge  lags  the  voltage,  Em  sin  (wt  +  a),  impressed 
across  the  circuit,  by  the  angle  0'.  The  angle  0',  therefore, 
represents  the  lag  of  the  charge  behind  the  impressed  voltage  or 
the  lead  of  the  impressed  voltage  with  respect  to  the  charge. 

When  —~  is  greater  than  coL,  tan  0'  is  positive  and  0'  is  less  than 
oou 


90  degrees.     When        is  less  than  o>L,  tan  0'  is  negative  and  the 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       165 


charge  lags  the  impressed  voltage  by  more  than  90  degrees  but 
less  than  180  degrees.  If  r  is  zero,  9'  is  also  zero.  In  this  case  the 
charge  is  in  phase  with  the  voltage  impressed  across  the  circuit, 
which,  under  this  condition,  is  the  same  as  the  voltage  across 
the  condenser. 

The  expression  for  the  charge  when  steady  conditions  have 
been  reached  is  therefore 


sin  <  ut  +  a  —  tan"1 


(140) 


The  angle  tf  =  tan"1 


in  equation  (140)  is  a  lag 


angle  since  equation  (140)  is  an  equation  for  the  charge  in  terms 
of  the  impressed  electromotive  force.  The  charge  lags  the 
impressed  electromotive  force  Em  sin  (wt  +  «)  by  the  angle  6'. 

Since 


-T.,  the  current  under  steady  conditions  is 


E 


cos 


sin 


at  +  a  —  tan"1 


at  +a 


—  tan" 


90°         (141) 


3 1  (uL~) 
u>G  ' 


FIG.  53. 


The  current,  therefore,  leads  the  charge  by  90  degrees.     The 
vector  representing  the  current  is  plotted  in  Fig.  53. 


166 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Let  6  be  the  angle  which  the  current  makes  with  the  impressed 
electromotive  force,  Em.     Then 

e  =  ef  -  90° 
i 

c  m 


—  6  =  tai 


=  —  tan- 


Equation  (141)  may  therefore  be  written 


+ 


Em 


sm 


cot  -ha  -  tan-1 1 

\ 


(142) 




The  signs  of  coL  and     ^  in  the  radical  A/r2  +  (coL ~J  ,  of 

equation  (142),  are  reversed  in  the  expression  (coL  —  — ^j  to 
make  it  correspond  to  the  like  term  in  the  expression  for  the  tan- 
gent of  the  angle  6.  This  can  be  done  since  the  term  (coL ^j 

is  squared  and  changing  its  sign  does  not  alter  its  squared 
value. 

The  complete  solution  of  equation  (130),  page  161,  for  charge 
is  given  by  adding  equation  (140)  to  equation  (131),  page  162. 


q  =  Y  +  u 


+  A*a* 
+  - 


A', 


(i  -  <*) 


sin 


ut  -j-  a 


/J_ 


COL) 


(143) 


The  complete  solution  of  equation  (130),  page  161  for  current, 
is  given  by  adding  equation  (142)  to  equation  (132),  page  162. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       167 

*  =  Y  +  u 


+  -          --         =  sin 


a-tan- 


(144) 


After  a  brief  interval  of  time,  the  transient  terms  in  equations 
(143)  and  (144)  become  sensibly  zero  for  circuits  with  constants 
ordinarily  met  in  practice  and  the  charge  and  current  then  be- 
come simple  harmonic  functions  of  time. 

The  maximum  value  of  the  current  under  steady  conditions  is 
found  by  dividing  the  maximum  value  of  the  impressed  electro- 

motive   force    by  -Jr2  +  ^coL  —  ^j  .     The    current    lags    the 

"L  ~  coC  1 

voltage  by  an  angle  whose  tangent  is  --  -  --  .     When  —Q  is 

greater  than  coL,  the  angle  6  becomes  negative  and  is  then  equiva- 
lent to  an  angle  of  lead.  In  this  case  the  current  actually  leads 
the  electromotive  force  impressed  on  the  circuit  by  an  angle  0. 

When  —  ~  is  less  than  o>L,  the  angle  6  is  positive  and  is  actually  an 

angle  of  lag.  In  this  case  the  current  actually  lags  the  voltage 
impressed  on  the  circuit. 

Since  the  effective  value  of  a  sinusoidal  wave  is  equal  to  its 
maximum  value  divided  by  the  square  root  of  two,  it  is  evident 
that  the  effective  value  of  the  current  is  found  by  dividing  the 


effective   value   of   the   voltage   by  A/r2  +  (o>L  -   —~\  .     This 

current  leads  or  lag 
greater  or  less  than 


current  leads  or  lags  the  impressed  voltage  according  as  —~  is 


E  E 


6  =  tan-1  I-      —  (146) 


168  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Impedance  and  Reactance. — The  quantity 

(147) 


is  called  the  impedance  and  is  measured  in  ohms.  It  is  constant 
only  when  resistance,  self -inductance,  capacitance  and  frequency 
are  constant.  It  may  increase  or  decrease  with  an  increase  in 

frequency  depending  on  the  relative  values  of  coL  and  -~.  The 
expression  coL  =  XL  is  the  inductive  reactance  and  the  expression 
pr  =  —  xc  is  the  capacitive  reactance.  (coL  —  —~\  =  XL  —  xc 

=  x0  is  the  resultant  reactance.  It  should  be  noted  that  in- 
ductive reactance  is  always  positive  while  capacitive  reactance 
is  always  negative.  The  resultant  reactance  may  be  either 
positive  or  negative  depending  upon  the  relative  values  of  XL  and 
xc.  Inductive  reactance  always  increases  with  increase  of 
frequency.  Capacitive  reactance  always  decreases  with  increase 
of  frequency. 

1  77F 

When  coL  =  — ^,  XQ  =  0  and  the  current  is  given  by  --  and 

U)O  T 

is  in  phase  with  the  impressed  electromotive  force.  Under  this 
condition  the  circuit  is  said  to  be  in  resonance.  More  will  be 
said  about  resonance  later. 

Vector  Method  of  Determining  the  Steady  Component  of  the 
Current  in  a  Circuit  Containing  Constant  Resistance,  Constant 
Self-inductance  and  Constant  Capacitance  in  Series. — The 
current  vector,  I  amperes  effective,  will  be  taken  along  the  axis 
of  reals.  The  effective  values  of  the  active  and  reactive  compo- 
nents of  the  impressed  electromotive  force  are  rl  and  J!(XL  —  xc) 

-  (  1  \ 

=  jl  ( coL  —  —~ )  respectively.  juLI  =  j!xL  is  the  reactive  com- 
ponent due  to  the  self-inductance.  ~J~r<  =  ~Jlxc  is  the  reactive 
component  due  to  the  capacitance.  jtaLI  leads  the  current  while 
— j— ~  lags  the  current. 

COL- 

E  =  r I  +  j(xL  -  xc)I  =  I[r+j(xL  -  xc)}  =  Yz      (148) 

When  substituting  numerical  values  it  must  be  remembered 
that  xc  is  negative. 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       169 

The  vectors  corresponding  to  equation  (148)  are  plotted  in 
Fig.  53,  page  165. 

The  complex  expression  for  the  impedance  is  therefore 

z  =  r  +  j(xL  —  xc) 

The  value  of  z  in  ohms  is  \/r2  +  (XL  -  xc)2>  Calling  the 
component  of  E  along  the  axis  of  reals,  i.e.,  the  real  or  active 
component,  Ea,  and  the  component  along  the  axis  of  imaginaries, 
i.e.,  the  imaginary  or  reactive  component,  ft,  gives 

E  =  Ea+jEr 
E  (volts)  =  \/Eaz  +  Er2 

(149) 


= 

z* 

r2  +  fe 
+  #. 

E,  -  ^c) 

s     '  J 

r2  + 

(XL 

-  XC) 

2 

(15C 

1 

a  a 

#«r  H 

-  ftfe 

-  XC) 

and  Ir 

= 

ftr- 

Ea 

(XL—XC) 

r2H 

-(%.- 

Xc)2 

r2  + 

(XL 

-XC)2 

. 
-  xc) 

Rationalizing  equation  (149),  to  get  rid  of  j  in  the  denominator, 
by  multiplying  the  numerator  and  the  denominator  each  by  the 
denominator  with  the  sign  of  the  term  involving  j  reversed,  gives 

=        Ea  +  jEr  r  -  j(xL  -  xc) 

*       r  +  j(xL  -  Xc)  '    r  -  j(xL  -  xc) 

Ear  +  Er(xL  -  xc)        .Err  -  Ea(xL  -  xc) 


where 


are  respectively  the  active  and  reactive  components  of  the 
current. 

But  /  is  along  the  axis  of  reals.     Therefore  Ir  =  0.     Hence 
Err  =  EO(XL  —  xc)  and 

ft  =  x^-^c  =  xo  = 
Ea  r  r 

where  0  is  the  angle  of  lag  of  the  current  with  respect  to  the 
voltage  impressed  on  the  circuit.  If  xc  is  greater  than  XL,  the 
angle  6  becomes  negative.  A  negative  angle  of  lag  is  equivalent 
to  a  positive  angle  of  lead.  In  this  case  the  current  actually 
leads  the  impressed  voltage.  If  xc  is  smaller  than  XL,  the  angle 
0  is  positive  and  is  actually  an  angle  of  lag.  In  this  case  the 
current  actually  lags  the  voltage  impressed  on  the  circuit.  The 
power  is  Eala  =  El  cos  0. 


170  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Polar  Expression  for  the  Impedance  of  a  Circuit  Containing 
Constant  Resistance,  Constant  Self-inductance  and  Constant 
Capacitance  in  Series.  —  The  polar  expression  for  the  impedance 
of  a  circuit  containing  constant  resistance,  constant  self-induc- 
tance and  constant  capacitance  in  series  is  (See  pages  136  and  150) 

z  =  z\e_  (151) 

where  the  angle  0  is  determined  by  the  relation  6  =  tan"1  — 

tan"1  —  .     If  the  inductive  reactance  predominates,  6  is  positive, 

while  if  the  capacitive  reactance  predominates,  6  is  negative. 
Impedance  may  be  written  in  four  different  ways,  namely: 

z  =  r  +  jxo 
=  z  (cos  6  +  j  sin  6) 


Equation  for  the  Velocity  and  Displacement  of  a  Mechanical 
System  Having  Friction,  Mass  and  Elasticity.  —  Before  leaving  the 
consideration  of  circuits  containing  constant  resistance,  constant 
self-inductance  and  constant  capacitance  in  series,  it  will  be  of 
interest  to  compare  the  equations  developed  for  such  circuits 
with  the  equations  for  displacement  and  velocity  of  a  mechanical 
system  having  constant  friction,  constant  mass  and  constant  elasti- 
city. It  has  already  been  pointed  out  that  self-inductance  and 
capacitance  are  respectively  analogues  of  mass  and  elasticity  in 
mechanics.  Electrical  resistance  corresponds  to  mechanical  fric- 
tion. Current  corresponds  to  velocity  and  charge  to  displace- 
ment in  a  mechanical  system. 

Consider  a  torsional  pendulum.  Let  c7  be  the  moment  of 
inertia  of  the  pendulum  about  its  axis  of  oscillation.  Also  let 
£  and  G  be  respectively  the  length  of  the  torsion  wire  and  its 
coefficient  of  torsional  rigidity.  The  radius  of  the  torsion  wire  is 
r.  Then  if  co  and  a  are  respectively  the  angular  velocity  and  the 
angular  displacement  of  the  pendulum  from  its  mean  position 
due  to  an  applied  couple  M  0 

Reaction  due  to  friction     =  M/  =  fc/co 

Reaction  due  to  inertia      =  Mi  —  ^y  — 

at 

7r7*^    fy 

Reaction  due  to  elasticity  =  Me  =  ff  ~n~  -^ 

&    L 


RESISTANCE,  INDUCTANCE  AND  CAPACITANCE       171 

Since  the  displacing  couple  MQ  must  balance  to  the  sum  of 
the  couples  due  to  the  reactions  caused  by  the  motion  of  the 
pendulum 

Mo  =  Mf  -f-  Mi  -f-  Me 


- 


_ 


- 


1 

_ 


da 
Since  co  =  -37  equation  (152)  may  be  written 


M< 


da 


. 

+ 


(152) 


(153) 


Equation  (153)  corresponds  exactly  to  equation  (90),  page  151, 
for  an  electrical  circuit  containing  constant  resistance,  constant 
self -inductance  and  constant  capacitance  in  series  and  its  solution 
takes  the  same  form  as  the  solution  for  that  equation. 

When  MQ  is  zero,  i.e.,  the  pendulum  oscillates  freely  without 
applied  accelerating  or  retarding  couples,  equation  (153)  becomes 

0  =  k^t +  kiw  + t '  (154) 

Equation  (154)  corresponds  to  equation  (117),  page  158,  for  the 
discharge  of  an  electrical  circuit  containing  constant  resistance, 
constant  self -inductance  and  constant  capacitance  in  series.  The 
solution  of  equation  (154)  takes  three  forms  corresponding  to  cases 
a,  6,  and  c  for  equation  (117). 


Case 


Electrical  circuit 


Mechanical  system 


(a) 


Charge  Q  is  a  maximum  when 
t  =  0.  It  then  decreases  ap- 
proaching zero  as  a  limit.  Cur- 
rent starts  at  zero  when  t  =  0, 
rises  to  a  maximum  then  de- 
creases approaching  zero  as  a 
limit.  There  is  no  oscillation. 


Displacement  a  is  a  maximum 
when  t  =  0.  It  then  decreases 
approaching  zero  as  a  limit.  An- 
gular velocity  co  starts  at  zero 
when  t  =  0,  rises  to  a  maximum 
then  decreases  approaching  zero 
as  a  limit.  There  is  no  oscilla- 
tion. 


172 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Case 


Electrical  circuit 


Mechanical  system 


Charge  and  current  have  the 
same  initial  and  final  values  as  in 
Case  (a).  They  oscillate  about 
these  values  with  an  amplitude 
which  decreases  logarithmically. 


Displacement  and  velocity  have 
the  same  initial  and  final  values  as 
in  Case  (a).  They  oscillate  about 
these  values  with  an  amplitude 
which  decreases  logarithmically. 


(c) 


r  =  ZA  /— 


2-K* 


Charge  and  current  have  the 
same  initial  and  final  values  as  in 
Cases  (a)  and  (6).  There  is  no 
oscillation. 


Displacement  and  velocity  have 
the  same  initial  and  final  values  as 
in  Cases  (a)  and  (6).  There  is  no 
oscillation. 


fe 
When  kf  <  2  */^  there  is  oscillation,  i.e.,  if  the  pendulum  is 

displaced  it  will  come  to  rest  after  a  series  of  oscillations  about  its 
mean  position.  These  oscillations  will  decrease  logarithmically 
with  time.  The  period  of  vibration  is 


When  kf  is  small, 


T  = 


T  = 


1  - 


ke 


Although  the  resistance  of  an  electrical  circuit  may  frequently 
be  too  high  for  electrical  oscillations  to  take  place,  the  friction  of 
a  pendulum  is  seldom  too  high  to  prevent  oscillation  unless  the 
pendulum  is  specially  damped.  The  vibrating  element  of  an 
oscillograph  is  an  example  of  a  highly  damped  torsional  pendu- 
lum. If  it  is  displaced,  it  will  come  to  rest  without  oscillation. 


CHAPTER  VI 

MUTUAL-INDUCTION 

Mutual-induction. — Up  to  the  present  point,  the  only  reactions 
which  have  been  considered,  when  a  current  in  a  circuit  is  varied, 
are  those  existing  or  arising  in  the  circuit  itself.  The  equations 
which  have  been  developed  hold  so  long  as  the  circuit  considered 
is  not  in  the  neighborhood  of  other  circuits  or  is  so  related  to 
any  other  circuit  in  its  vicinity  that  there  can  be  no  interaction 
between  them.  In  general,  however,  when  two  or  more  circuits 
are  in  proximity,  any  change  in  the  current  in  one  of  them  will 
cause  inductive  effects  in  the  others. 

Consider  two  circuits  which  are  in  proximity  to  each  other. 
When  one  of  two  such  circuits  carries  current  a  magnetic  field  will 
be  established.  This  field  will  link  not  only  the  circuit  by  which 
it  is  produced  but,  in  general,  a  certain  portion  of  it  will  also 
link  the  other  circuit.  The  relative  amounts  of  flux  linking  each 
circuit  will  depend  chiefly  on  the  relative  size  and  shape  of  the 
circuits,  their  relative  position  and  the  permeability  of  the  sur- 
rounding medium.  All  of  the  flux  produced  by  one  circuit  can 
never  link  the  other,  although  the  difference  between  the  flux 
linking  one  and  that  linking  the  other  may  be  made  very  small 
by  closely  interwinding  them.  On  the  other  hand,  the  difference 
may  be  very  great  when  the  circuits  are  far  removed  from  each 
other  or  are  placed  with  the  axis  of  one  in  the  plane  of  the  other. 

If  each  circuit  is  carrying  current,  any  change  in  the  current 
of  either  will  be  accompanied  by  a  change  in  the  flux  linkages  of 
each  circuit.  The  change  in  the  flux  linkages  of  each  circuit  will 
induce  in  each  a  voltage  which  will  be  equal  to  the  time  rate  of 
change  of  flux  linkages  for  the  circuit.  The  rate  of  change  of  flux 
linkages  of  one  circuit  with  respect  to  change  in  current  in  the 
other  forms  one  of  the  most  important  and  fundamental  constants 
in  the  electric  theory  of  circuits. 

If  both  circuits  are  now  closed  and  carrying  current,  any 
change  in  the  current  in  one  will  induce  in  the  other  a  voltage 

173 


174  PRINCIPLES  OF  ALTERNATING  CURRENTS 

which  will  cause  a  current  and,  according  to  the  law  of  Lenz, 
this  second  current  must  have  such  a  direction  as  to  oppose 
the  change  in  the  flux  producing  it.  When  the  current  in  the 
first  circuit  increases,  the  current  induced  in  the  second  circuit 
must  be  in  a  direction  opposite  to  that  in  the  first  circuit.  When 
the  current  in  the  first  circuit  decreases,  the  current  induced  in 
the  second  circuit  must  be  in  the  same  direction  as  that  in  the  first 
circuit.  No  change  can  take  place  in  the  current  of  one  without 
producing  a  corresponding  change  by  induction  in  the  current  of 
the  other.  Any  change  in  the  current  of  one  will  result  in  a  cur- 
rent in  the  other  as  there  is  mutual-induction  between  two. 

Mutual-induction  plays  an  important  part  in  the  operation 
of  most  alternating-current  apparatus.  The  operation  of  certain 
types  of  apparatus,  such  as  the  transformer  and  the  induction  mo- 
tor, depends  entirely  upon  it.  Without  the  transformer  and  the 
induction  motor,  the  present  development  of  alternating-current 
systems  of  power  distribution  and  utilization  would  be  impossible. 

Coefficient  of  Mutual-induction  or  Mutual-inductance. — The 
self-inductance  of  a  circuit  or  its  coefficient  of  self-induction 
was  defined  as  the  rate  of  change  in  the  flux  linkages  of  the  circuit 
with  respect  to  its  current.  When  the  permeability  of  the  sur- 
rounding medium  is  constant,  the  self-inductance  of  the  circuit 
is  constant.  When  the  permeability  varies  with  current  strength, 
the  self -inductance  also  varies  with  current  strength.  When 
the  permeability  is  constant,  self-inductance  may  be  defined 
as  the  change  in  the  flux  linkages  of  the  circuit  per  unit  change  in 
its  current.  These  definitions  of  self-inductance  assume  that  no 
change  is  produced  in  the  currents  in  other  circuits  in  the  neigh- 
borhood. When  the  current  is  varied  in  a  circuit  having  a  self- 
inductance  L,  a  voltage  of  self-induction  is  induced  in  it  which  is 
equal  to  the  time  rate  of  change  of  the  flux  linkages  of  the  circuit. 
The  voltage  drop  due  to  self-inductance  (assumed  constant)  is 

fti  =  L^  (1) 

Similarly,  the  mutual-inductance  or  coefficient  of  mutual- 
induction  of  two  circuits  is  defined  as  the  rate  of  change  in 
the  flux  linkages  of  one  with  respect  to  current  in  the  other. 
When  the  permeability  of  the  circuits  is  constant,  their  mutual- 
inductance  is  constant.  When  the  permeability  varies  with  the 


MUTUAL-INDUCTION  175 

current  strength  in  either  circuit,  the  mutual-inductance  also 
varies  with  current  strength.  When  the  permeability  is  constant, 
the  mutual-inductance  of  the  two  circuits  may  be  defined  as  the 
change  of  flux  linkages  for  either  per  unit  change  of  current  in  the 
other.  When  the  current  is  varied  in  one  of  the  two  circuits 
having  a  mutual-inductance  M  ,  a  voltage  of  mutual-induction  is 
induced  in  the  other  and  is  equal  to  the  time  rate  of  change  of 
the  flux  linkages  for  that  circuit.  The  voltage  drop  in  circuit  2 
due  to  the  mutual-inductance  (assumed  constant)  of  circuit  1 
on  circuit  2  is 

c«2  =  Mn^  (2) 

Self-inductance  and  mutual-inductance  are  both  measured  in 
the  same  unit,  the  henry.  The  mutual-inductance  of  two 
circuits  is  one  henry  when  the  rate  of  change  of  flux  linkages 
of  either,  with  respect  to  the  current  in  amperes  in  the  other,  is 
108  flux  linkages  per  ampere.  When  the  permeability  is  constant, 
two  circuits  have  a  mutual-inductance  of  one  henry  when  a  change 
of  108  flux  linkages  is  produced  in  either  by  a  change  of  one  am- 
pere in  the  current  of  the  other.  These  definitions  assume  that 
the  current  in  one  circuit  is  constant  while  the  current  in  the 
other  circuit  is  changed.  When  the  coefficient  of  self-  or  mutual- 
induction  is  expressed  in  henrys  and  the  rate  of  change  of  current 
is  in  amperes  per  second,  the  induced  voltage  is  in  volts. 

If  the  permeability  of  a  circuit  is  constant  and  there  is  no 
magnetic  leakage  between  its  turns,  i.e.,  if  all  the  flux  produced 
by  each  turn  links  all  the  turns,  the  coefficient  of  self-induction  is 


Ll  =       T^1  abhenrys  (3) 

In 

where  Ni  is  the  number  of  turns  and  (R  is  the  reluctance  of  the 
magnetic  circuit. 

If  N2  is  the  number  of  turns  in  a  second  circuit  so  related 
to  the  first  that  all  the  flux  produced  by  the  first  circuit  links 
all  of  the  turns  of  the  second  circuit,  the  coefficient  of  mutual- 
induction  of  the  first  circuit  on  the  second  circuit  will  be 

Mi2  =  -    —  N2  abhenrys.  (4) 

01 

All  the  flux  produced  by  the  first  circuit  can  never  link  all  the 
turns  of  the  second  circuit,  and  all  the  flux  produced  by  the 


176  PRINCIPLES  OF  ALTERNATING  CURRENTS 

second  circuit  can  never  link  all  the  turns  of  the  first  circuit. 
There  will  always  be  some  leakage  of  magnetic  flux  between  the 
two  circuits.  By  closely  interwinding  the  two  circuits,  the 
magnetic  leakage  may  be  made  very  small.  It  is,  however,  often 
very  great. 

Although  it  is  possible  to  calculate  the  self-inductance  and 
mutual-inductance  of  certain  very  simple  circuits  with  a  fair 
degree  of  accuracy,  accurate  calculation  of  either  self-inductance 
or  mutual-inductance  is  usually  impossible. 

Voltage  Drop  across  Circuits  Having  Resistance  and  Self-  and 
Mutual-inductance. — From  the  definitions  of  self-  and  mutual- 
inductance,  it  is  obvious  that  the  voltage  drops  across  two  circuits 
having  resistance  and  self-  and  mutual-inductance  are  given  by 
the  following  differential  equations. 

t>!  =  nt !  +  L,—1  +  M21^2  (5) 

v,  =  r«*  +  L2^2  +  M^l  (6) 

When  no  magnetic  material  is  present,  the  coefficients  of 
self-  and  mutual-induction  in  equations  (5)  and  (6)  are  constant. 
When  the  flux  produced  by  the  currents  i\  and  i%  is  in  magnetic 
material,  the  coefficients  of  self-  and  mutual-induction  will 
vary  with  the  currents.  They  will  be  complicated  functions  of 
the  currents.  In  general,  when  magnetic  material  is  present, 
no  exact  solution  of  equations  (5)  and  (6)  can  be  obtained. 

It  will  be  shown  that  the  coefficient  of  mutual-induction 
of  two  circuits,  in  the  absence  of  magnetic  material,  is  the  same 
whether  it  is  taken  for  circuit  1  with  respect  to  circuit  2  or  for 
circuit  2  with  respect  to  circuit  1.  Under  this  condition  the 
subscripts  on  the  coefficient  of  mutual-induction,  M,  have  no 
significance  and  may  be  omitted. 

The  Coefficients  of  Mutual-induction,  Mi2  and  M2i,  of  Two 
Circuits  in  a  Medium  of  Constant  Permeability  are  Equal. — Let 
the  permeability  of  the  medium  be  constant.  This  assumes  that 
there  is  no  magnetic  material  present.  The  coefficients  of 
mutual-induction  M i2,  of  circuit  1  on  circuit  2,  and  M2i,  of  cir- 
cuit 2  on  circuit  1,  are  equal.  This  statement  is  true  without 
regard  to  the  size,  shape  or  position  of  the  two  circuits. 


MUTUAL-INDUCTION  177 

The  total  electromagnetic  energy  in  an  electrical  system 
consisting  of  two  circuits  1  and  2,  which  have  constant  self-  and 
mutual-inductance  and  carry  currents  I\  and  /2  respectively,  is 
obviously  independent  of  the  order  or  manner  in  which  the  cur- 
rents are  established.  This  follows  from  the  law  of  conservation 
of  energy. 

Let  1 1  and  1 2  be  the  final  values  of  the  currents  in  the  circuits 
1  and  2  respectively.  Consider  72  to  be  zero  and  establish 
1 1.  The  energy  due  to  establishing  /i  will  be  due  entirely  to  the 
self -inductance  of  circuit  1.  While  I\  is  increasing,  a  voltage 
will  be  induced  in  circuit  2  by  the  mutual-induction  of  circuit  1 
on  circuit  2.  No  work  can  be  done  in  circuit  2  by  this  voltage, 
since  1 2  is  zero.  The  total  electromagnetic  energy  due  to  estab- 
lishing the  current  7i  is 


(7) 

Now  consider  the  current  /i  to  be  maintained  constant  while 
the  current  1 2  is  established.  The  change  produced  in  the 
electromagnetic  energy  of  the  system  by  the  establishment  of  72 
will  be  made  up  of  two  parts :  one  due  to  the  effect  on  1 2  of  the 
voltage  induced  in  circuit  2,  the  other,  due  to  the  effect  on  I\  of 
the  voltage  induced  in  circuit  1.  The  first  part  is  due  to  the 
self-inductance  of  circuit  2,  while  the  second  part  is  due  to  the 
mutual-inductance  of  circuit  2  on  circuit  1.  The  change  pro- 
duced in  the  electromagnetic  energy  of  the  system  by  the  estab- 
lishment of  72  is 


=    L2/22  +  Jlf  will,  (8) 

The  total  change  produced  in  the  electromagnetic  energy 
of  the  system  by  the  establishment  of  the  currents  /i  and  72  is 
therefore 

W0  =  Wl+W2  =  1LJS  +  il,2/22  +  M21/i/a  (9) 


178 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


If  72  had  been  established  first,  the  energy  equation  would 
have  been 


W0'  = 


W  i  = 


(10) 


Since,  according  to  the  law  of  the  conservation  of  energy, 
WQ  and  Wd  must  be  equal,  Mi2  and  M2i  must  also  be  equal.  In 
other  words,  the  mutual-inductance  of  circuit  1  with  respect  to 
circuit  2  is  equal  to  the  mutual-inductance  of  circuit  2  with 
respect  to  circuit  1.  In  general,  when  there  is  no  magnetic 
material  present,  the  coefficient  of  mutual-induction  of  two 
circuits  is  the  same  whether  it  is  considered  with  respect  to 
circuit  1  on  circuit  2  or  with  respect  to  circuit  2  on  circuit  1. 


FIG.  54. 

Magnetic  Leakage  and  Leakage  Coefficients. — Fig.  54  shows 
two  circuits  having  self-  and  mutual-inductance.  All  the  flux 
that  links  circuit  1  does  not  link  circuit  2,  and  all  the  flux  that 
links  circuit  2  does  not  link  circuit  1.  The  flux  that  is  common 
to  both  circuits,  i.e.,  the  mutual  flux,  and  the  fluxes  that  link 
one  circuit  without  linking  the  other,  i.e.,  the  leakage  fluxes,  are 
indicated. 

In  Fig.  54,  <pM  is  the  mutual  flux.  <psi  and  <^S2  are  the  leakage 
fluxes  for  circuits  1  and  2  respectively. 

Let  Li  and  L2  be  the  self -inductance  of  the  circuits  1  and  2, 
whose  turns  are  Ni  and  N%  respectively.  Assume  there  is  no 
magnetic  material  present.  Under  this  condition  the  self- 
inductances,  LI  and  L2,  will  be  constant.  Then  if  all  the  flux 


MUTUAL-INDUCTION  179 

produced  in  each  turn  of  either  circuit  links  all  the  turns  of  that 
circuit 


L,  =  (ID 

r 


where  (R  is  the  reluctance  of  the  magnetic  circuit. 

Equations  (11)  and  (12)  would  be  approximately  correct 
for  solenoids  which  were  long  compared  with  their  diameters. 

It  is  obvious  that  all  the  flux  produced  by  either  circuit  cannot 
link  all  the  turns  of  the  other.  There  must  always  be  some 
magnetic  leakage  between  the  two  circuits  although,  in  certain 
cases,  the  leakage  maybe  very  small.  On  the  other  hand,  it  may 
be  made  very  large  by  increasing  the  separation  of  the  circuits 
and  suitably  changing  their  relative  positions. 

The  magnetic  leakage  between  two  circuits  is  least  when  they 
are  interwound,  with  the  turns  of  one  in  close  proximity  to  the 
turns  of  the  other.  The  leakage  would  be  zero  if  it  were  possible 
to  make  the  corresponding  turns  of  the  two  circuits  occupy 
exactly  the  same  position.  The  magnetic  leakage  is  greatest 
when  the  two  circuits  are  far  apart  and  are  placed  so  that  the 
axis  of  one  is  hi  the  plane  of  the  other.  In  the  latter  case  no 
flux  produced  by  either  circuit  would  link  the  other  and  the 
mutual-inductance  of  the  circuits  would  be  zero. 

The  self-inductance  of  a  circuit  depends  upon  its  size,  shape, 
number  of  turns  and  reluctance  of  its  magnetic  circuit.  The 
mutual-inductance  of  two  circuits  depends  on  their  size,  shape, 
number  of  turns,  relative  position  and  reluctance  of  the  mag- 
netic circuit.  Magnetic  leakage  is  determined  by  the  relative 
size,  shape  and  position  of  the  circuits,  the  way  the  circuits  are 
wound,  i.e.,  whether  compact  or  spread  out,  and,  if  magnetic 
material  is  present,  on  the  degree  of  saturation  of  the  mag- 
netic circuit.  When  an  iron  core  is  used,  the  leakage  will 
depend  very  largely  on  the  position  of  the  circuits  on  the  core. 

Let  <p\  be  the  total  flux  linking  circuit  1  when  it  carries  a 
current  /i,  the  current  72  in  circuit  2  being  zero.  Let  .<pMi  be  the 
portion  of  <p\  linking  circuit  2.  Then 

=  fc,  (13) 


180  PRINCIPLES  OF  ALTERNATING  CURRENTS 

is  known  as  the  leakage  coefficient  of  circuit  1  with  respect  to 
circuit  2. 

Similarly,  if  <p2  is  the  total  flux  linking  circuit  2  when  it  carries 
a  current  /2,  the  current  I\  being  zero,  and  <pMz  is  the  portion  of 
p2  linking  circuit  1, 

=  k*  (14) 


is  the  leakage  coefficient  of  circuit  2  with  respect  to  circuit  1. 

The  leakage  coefficients  of  two  circuits  are  not  necessarily 
constant  and  independent  of  the  current,  and  they  are  not 
necessarily  equal.  If  the  two  circuits  are  of  exactly  the  same 
shape  and  size  and  are  symmetrically  placed,  their  leakage 
coefficients  will  be  equal.  The  leakage  coefficients  will  be 
constant  and  independent  of  the  current  when  there  is  no 
magnetic  material  in  the  path  of  the  fluxes.  They  may  be  mate- 
rially altered  by  the  presence  of  magnetic  material  in  the  path 
of  the  mutual  flux. 

If  ki  and  &2  are  the  leakage  coefficients  of  two  circuits  having 
mutual-  and  self-inductance 


(16) 

It  has  already  been  shown  that 

MIZ  =  MM  =  M 

when  the  permeability  of  the  surrounding  medium  is  constant. 
Relation  among  the  Mutual-  and  Self-inductances  of  Two 
Circuits  Containing  No  Magnetic  Material,  i.e.,  Having  Constant 
Magnetic  Reluctance. — Since  Mi2  and  M^i  are  equal,  it  follows 
from  equations  (15)  and  (16)  that 


M  =  M 12  =  M21  =  VLi(l  -  fci)  XL2(l  -fc2)        (17) 

If  there  were  no  leakage,  i.e.,  if  ki  and  &2  were  both  zero,  the 
mutual  inductance  of  two  circuits,  whose  self-inductances  are 
LI  and  L2,  would  be  equal  to  \/Z/i  X  L2. 

Coefficient  of  Electromagnetic  Coupling  between  Two  Electric 
Circuits  Having  Mutual-inductance. — The  coefficient  of  electro- 


MUTUAL-INDUCTION 


181 


magnetic  coupling  between  two  circuits  having  mutual  induc- 
tance is  the  ratio  of  M  to  \/£>i  X  L2. 


Coefficient  of  coupling  = 


M 


i  X  L 


(18) 


-  fci)  X 


V(i  -  fci)  x  (i  - 


(19) 


The  coupling  between  two  circuits  is  said  to  be  close  when 
the  coefficient  of  coupling  is  large.  In  radio  work,  circuits 
which  have  a  coupling  greater  than  0.5  are  said  to  be  close 
coupled.  The  degree  of  coupling  that  is  desirable  depends  on 
the  purpose  for  which  the  circuits  are  to  be  used.  For  commer- 
cial transformers  with  iron  cores  the  coupling  between  the 
primary  and  secondary  windings  is  very  close  and  may  be  as 
high  as  0.98  or  0.99.  Close  coupling  is  generally  desired  in  a 
commercial  transformer,  since  the  voltage  regulation  of  a  trans- 
former depends  very  largely  on  the  closeness  of  coupling  between 
its  primary  and  secondary  windings.  Good  voltage  regulation 
requires  small  magnetic  leakage  between  primary  and  secondary 
windings  and  therefore  close  coupling.  Such  close  coupling  as 
is  used  in  commercial  power 
transformers  is  undesirable  in 
transformers  for  radio  work. 

General  Equations  for  the 
Voltage  Drops  Across  Two  In- 
ductively Coupled  Circuits  Each 
Having  Constant  Resistance, 
Constant  Self -inductance  and 
Constant  Capacitance  in  Series.— 
Let  r,  L  and  C  with  subscripts 
1  and  2  be  the  constants  of  the  two  coupled  circuits  and  let  M 
be  their  mutual-inductance.  The  diagram  of  connections  is 
given  in  Fig.  55. 

If  vi  and  vz  are  the  instantaneous  voltage  drops  across  the 
two  circuits  when  their  currents  are  i\  and  t"2,  the  following  differ- 
ential equations  hold 


FIG.  55. 


182  PRINCIPLES  OF  ALTERNATING  CURRENTS 

vi  =  riii  +  L^  +  i  I  i^t  +  M^  (20) 

Cit          C  i,/  Clt 

(21) 

Equations  (20)  and  (21)  are  the  fundamental  differential 
equations  for  inductively  coupled  circuits  such  are  used  for 
radio-telegraphy. 

A  simple  and  important  case  of  the  solution  of  equations 
(20)  and  (21)  occurs  when  the  circuits  have  only  resistance  and 
self-  and  mutual-inductance  and  the  second  circuit  is  short-cir- 
cuited. This  is  equivalent  to  saying  that  the  series  capacitances 
Ci  and  Cz  are  infinite.  In  this  case  equation  (21)  becomes 

diz  _         (  M  dii      Trti 
~dt   ~~         \Lz~di      ~Lz 

Substituting  the  value  of  -rr  from  equation  (22)  in  equation 
(20)  and  remembering  that  Ci  is  infinite 

v    =  r  i   +  I  L    -  —  \^l——ri  (23) 

1  L2  /  dt       Lz  2' 

If  TZ  is  small,  equation  (23)  may  be  written 

PI  =  riii  +   s  LI  -    -=-  >  -7,-,  approximately.  (24) 

[  LZ  )   dt 

Equation  (24)  is  of  the  same  form  as  equation  (1),  page  120, 
for  a  simple  series  circuit  containing  resistance  and  self-induc- 

f  M2 } 

tance  in  series  except  that  \  LI  —  -j —  }  replaces  LI. 

I  Liz  ) 

The  effect  of  the  short-circuited  coupled  circuit  is  to  diminish 
the  apparent  self -inductance  of  the  primary  circuit.  If  the  coeffi- 
cient of  coupling  could  be  made  unity,  y-  would  be  equal  to  LI. 

f          M2 } 
In  this  case  the  apparent  inductance,  \  LI  —  -y-  M  of  the  primary 

circuit  would  be  zero  and  it  would  act  like  a  circuit  containing 
only  pure  resistance.     Although  it  is  not  possible  to  make  the 


MUTUAL-INDUCTION  183 

apparent  inductance  of  the  primary  circuit  exactly  zero,  it  may 
be  made  very  low  by  interwinding  the  two  circuits  so  as  to  make 
the  magnetic  leakage  between  them  very  small. 

A  closely  coupled,  low  resistance,  short-circuited  secondary 
winding  may  be  used  to  suppress  the  arc  when  a  relay  circuit, 
which  carries  direct  current,  is  opened.  For  this  purpose  the 
main  relay  winding  would  probably  be  wound  on  a  copper  cylin- 
der which  would  serve  as  the  low  resistance,  closely  coupled, 
secondary  winding.  This  cylinder  would  have  no  effect  on  the 
operation  of  the  relay  so  long  as  the  exciting  current  was  steady. 
When  the  exciting  current  varied,  however,  the  current  induced 
in  the  low  resistance,  closely  coupled  winding  would  decrease  the 
apparent  inductance  of  the  exciting  winding.  The  presence  of 
the  coupled  winding  would  cause  the  exciting  current  to  reach 
its  final  steady  value  more  quickly  after  the  circuit  was  closed. 
By  decreasing  the  apparent  inductance  of  the  exciting  winding, 
it  would  also  practically  suppress  the  arc  when  the  exciting 
circuit  was  broken.  The  effect  of  the  coupled  winding  is  to  de- 
crease the  apparent  time  constant  of  the  primary  winding. 

Leakage -inductance  of  Coupled  Circuits. — When  dealing 
with  inductively  coupled  circuits,  it  is  often  desirable  to  split 
the  self-inductance  of  each  winding  into  two  parts.  One  part, 
called  the  leakage  inductance,  is  produced  by  the  leakage  flux. 
The  other  part  is  due  to  that  portion  of  the  total  flux  of  self- 
induction  which  links  both  windings.  The  separation  of  self- 
inductance  into  these  two  parts  often  much  simplifies  problems 
involving  inductively  coupled  circuits.  The  treatment  of  the 
alternating-current  transformer  and  also  of  the  induction  motor 
is  very  much  simplified  by  the  use  of  this  device. 

Self-inductance,  L,  has  been  defined  as  the  change  in  the 
flux  linkages  ef  a  circuit  per  unit  change  in  its  current.  The 
flux  concerned  includes  all  of  the  flux  produced  by  the  current 
when  it  acts  alone.  Similarly,  the  mutual-inductance,  M,  of 
two  circuits  has  been  defined  as  the  change  produced  in  the  flux 
linkages  of  one  circuit  by  unit  change  of  current  in  the  other. 
In  this  case  the  flux  concerned  includes  only  that  portion  of  the 
total  flux  which  links  both  windings  when  one  alone  carries  current. 
It  does  not  include  any  leakage  flux,  since  the  leakage  flux,  by 
its  definition,  cannot  link  the  second  circuit.  Leakage-indue- 


184  PRINCIPLES  OF  ALTERNATING  CURRENTS 

tance  is  defined  in  a  similar  manner.  The  leakage-inductance, 
Sj  of  a  circuit  is  the  rate  of  change  of  the  leakage-flux  linkages 
per  unit  current.  The  flux  concerned  in  leakage-inductance 
includes  only  that  portion  of  the  total  flux  which  does  not  link 
the  second  circuit.  Leakage-inductance,  as  well  as  self-  and 
mutual-inductance,  is  measured  in  henrys.  The  leakage-in- 
ductance of  a  circuit  is  one  henry  when  a  change  of  one  ampere 
in  its  current  causes  a  change  of  108  leakage-flux  linkages.  The 
leakage-inductance  of  a  circuit  in  henrys  multiplied  by  27r/, 
where  /  is  the  frequency,  is  its  leakage  reactance  in  ohms.  Just 

as  eiL  =  LI-JJ  is  the  voltage  drop  produced  in  circuit  1  by  self- 
inductance,  and  CZM  =  M-TJ  is  the  voltage  drop  produced  in 

circuit  2  by  mutual-inductance,  so  eis  =  $1777  is  the  voltage  drop 

produced  in  circuit  1  by  leakage-inductance.  The  corresponding 
root-mean-square  or  effective  values  of  the  voltages  are 

Eu,  =  2n/Li/i  =  coLJi, 

E*M    =    27T/MI,    =    uMIi, 
EIS      =    2firfSlIi    =    C0$i/i. 

If  the  inductances  are  expressed  in  henrys,  the  currents  in 
amperes  and  the  frequency  in  cycles  per  second,  the  voltages 
will  be  in  volts. 

Let  (pis  and  (p2S  be  the  leakage  fluxes  for  circuits  1  and  2. 
Assuming  that  all  of  the  leakage  flux  links  all  of  the  turns  of  the 
circuit  in  which  it  is  produced,  the  leakage  inductances  are 

S,  =  Ni  ^  10-8    henrys  (25) 

ai\ 

S2=N2d^  10-8  henrys  (26) 

6^2 

where  the  currents  are  expressed  in  amperes. 
If  the  leakage  fluxes  per  ampere  are  constant 

S,  =  N^~  10~8  henrys  (27) 

^i 

Si  =  N2  ^  10-8  henrys  (28) 


MUTUAL-INDUCTION  185 

Relations  Among  the  Fluxes  Corresponding  to  Self  -inductance, 
Leakage  -inductance  and  Mutual-inductance.  —  Consider  two 
coupled  circuits  having  N\  and  Nz  turns.  Let  the  inductances  be 
expressed  in  henrys.  Assume  the  inductances  to  be  constant. 
This  is  equivalent  to  saying  that  there  is  no  magnetic  material  in 
the  vicinity  of  the  circuits.  For  7\  amperes  in  circuit  1,  circuit  2 
being  open, 

~  1  1  X  108  =   (p\  maxwells  =  total  flux  linking  circuit  1  due  to  its 
^1  self  -inductance. 

Cf 

-j^-  1  1  X  108  =  <PLS  maxwells  =  leakage  flux  linking  circuit  1  due  to 
*  l  its  leakage-inductance. 

•jT=r-  1  1  X  108  =  <PM  2  maxwells  =  flux  linking  circuit  2  due  to  the 
•^2  mutual-inductance  of  circuit  1  on 

circuit  2. 

For  /2  amperes  in  circuit  2,  circuit  1  being  open 

-—-  1  2  X  108  =    <f>2  maxwells  =  total  flux  linking  circuit  2  due  to 

its  self  -inductance. 

o 

j~  J2  X  108  =  (pzs  maxwells  =  leakage  flux  linking  circuit  2  due 
^2  to  its  leakage-inductance. 

-=;-  Iz  X  108  =  (pM\  maxwells  =  flux  linking  circuit  1  due  to  the 

mutual-inductance  of  circuit  2  on 
circuit  1. 
Obviously, 

<f>i  —  <Pis  =  <PM2  maxwells  =  part  of  <pi  which  also  links  circuit  2. 
<^2  —  <Pzs  =  <f>Mi  maxwells  =  part  of  <pz  which  also  links  circuit  1. 

From  the  preceding  equations  it  follows  that 


-  Si)  ~  X  108  =  JJT-  /!  X  108  =  <pM2  maxweUs. 


(Lt  -  Si)    ±  X  108  =        J2  X  108  =  VMI  maxwells 


(29) 


(so) 


186  PRINCIPLES  OF  ALTERNATING  CURRENTS 

From  equations  (29)  and  (30) 

(Li  -  SJ  X  (L2  -  S2)  =  M2  (31) 

and  

M  =  vUi  -  Si)  X  (L2  -  S2)  (32) 

But    since    LI  —  Si  =  I/i(l  —  ki)    and    L2  —  &  =  I/2(l  ~~  ^2) 
equation  (32)  is  equivalent  to  equation  ^17),  page  180. 

Now  let  circuits  1  and  2  carry  Ji  and  72  amperes,  respectively, 
at  the  same  time.  In  general,  the  currents  will  not  be  in  phase 
and  the  fluxes  they  produce  will  not  be  in  phase.  They  must  be 
added  vectorially .  Both  the  currents  and  fluxes  must  be  ex- 
pressed as  vectors. 

''^  +  ^1^   -  (33) 


M 

=  —  7-  {/2  Nz  +  7i  TV^  108  maxwells  (34) 

J\  ll\  2 

is  the  resultant  mutual  flux  linking  circuits  1  and  2  when  they 
carry  currents  Ii  and  72  respectively.  ^^^  is  the  vector  sum  of 
the  two  component  mutual  fluxes  <pMi  and  <pMz  produced  by  the 
currents  l\  and  72  respectively.  (pMR  is  the  maximum  or  the 
root-mean-square  value  of  the  resultant  mutual  flux,  according 
as  the  maximum  or  the  root-mean-square  values  of  the  currents 
are  used. 

When  both  circuits  carry  current,  it  is  seen  that  the  resultant 
mutual  flux  is  made  up  of  two  components:  one,  <pM\,  produced  by 
and  in  phase  with  Ii,  the  other,  <pMz,  produced  by  and  in  phase 
with  72.  The  resultant  ampere-turns  acting  on  the  magnetic 
circuit  to  produce  the  resultant  mutual  flux  <PMR  are  (1  2^2  + 
IiNi).  (See  equation  (34)) 


9  MR  <PlS    =    <PlR 

is  the  resultant  flux  linking  circuit  1  when  both  circuits  carry 
current.  It  is  the  vector  sum  of  the  resultant  mutual  flux  and 
the  leakage  flux  of  circuit  1.  The  total  voltage  induced  in 
circuit  1  is  produced  by  this  flux.  This  voltage  consists  of 
two  components,  one  due  to  the  leakage  flux  pis,  the  other  due 
to  the  resultant  mutual  flux  <pMR. 

<f>MR   -f"   <f>2S    =    <P2R 


MUTUAL-INDUCTION 


187 


is  the  resultant  flux  linking  circuit  2  when  both  circuits  carry 
current.  It  is  the  vector  sum  of  the  resultant  mutual  flux  and 
the  leakage  flux  of  circuit  2.  The  total  voltage  induced  in 
circuit  2  is  produced  by  this  flux.  This  voltage  consists  of 
two  components,  one  due  to  the  leakage  flux  <p2s,  the  other  due 
to  the  resultant  mutual  flux  <pMR. 

Voltages  Induced  in  the  Windings  of  an  Air-core  Trans- 
former.— Fig.  56  shows  a  transformer  with  two  coils  wound  on  a 
core. 

For  the   present   purpose   the   core  will   be  assumed  to  be 
of   non-magnetic  material.     When 
an    iron    core    is    used,    the    co- 
efficients   of  self-  and  mutual-in- 
duction    will    not     be     constant. 
They  will  vary  with  the  saturation 
of  the  magnetic  circuit.     However, 
the  presence  of  an  iron  core  in  com- 
mercial  alternating-current  trans- 
formers   does    not    introduce    any 
serious  difficulty,  if  the  self-induct- 
ances of  the  primary  and  secondary  windings  are  split  up  into 
two  parts,  one  due  to  leakage  flux  and  the  other  due  to  mutual 
flux. 

The  total  voltage  drop  in  the  primary  winding  due  to  self- 
and  mutual-inductance,  when  both  primary  and  secondary 
windings  carry  current,  is, 


Core 

T 

It 

Primary   ' 

„  • 


< 

—  - 

'Secondary 

Winding  < 
1 

„  

' 

— 

,  Winding 

FIG.  56. 


+ 


^Ni 


+  /       (liNi  +  hN2] 


(37) 


(38) 


Since  I\  and  7  2  are  not  in  phase,  equations  (37)  and  (38)  and 
the  equations  which  follow  are  true  only  when  considered  in  a 
vector  sense. 


i 
——  is  the  mutual-flux  linkages  with  circuit  1  due  to 

is    the    mutual-flux   linkages  with  circuit   1    due   to 


188  PRINCIPLES  OF  ALTERNATING  CURRENTS 

j^~(IiNi  +  72Af2)  is,  therefore,  the  resultant  mutual-flux  linkages 

with  circuit  1  due  the  combined  action  of  currents  7i  and  72. 
Dividing  this  by  NI  gives  the  resultant  mutual  flux  <PMR  due  to 
the  combined  action  of  currents  7i  and  72. 

M    /_ 

(39) 


The  first  term,  jwSJi,  of  the  second  member  of  equation 
(38)  ,  is  the  voltage  drop  produced  in  the  primary  winding  by  the 
primary  leakage  flux,  i.e.,  by  that  portion  of  the  total  primary 
flux  which  does  not  link  the  secondary  winding.  The  second 

term,  j^r(IiNi  +  72Ar2),  is  the  voltage  drop  produced  in  the 

1\  2 

primary  winding  by  the  resultant  mutual  flux,  <PMR,  due  to  the 

combined  action  of  the  primary  and  secondary  currents,  7i  and  72. 

The  total  voltage  drop  due  to  induction  in  the  secondary  wind- 

ing is  given  by  an  expression  exactly  similar  to  equation  (38). 

#2  =  j<o£272  +  /~  (l*N2  +  Ltf  ,)  (40) 

Since  power  is  absorbed  by  the  primary  winding  and  is  given 
out  by  the  secondary  winding,  the  primary  and  secondary  cur- 
rents, 1  1  and  72,  must  be,  in  a  general  sense,  in  phase  opposition 
and  must  therefore  produce  component  fluxes  which  are  also  in  a 
general  sense  in  phase  opposition.  The  current  7i  may  be 
considered  to  be  made  up  of  two  components,  one  of  which  is 

-  N<i 
equal  to  7  2^-  and  in  phase  opposition  to  72.     Call  this  component 

7/.     Call  the  other  component  7^,.     Then 

Jl'tfl    =     -/2#2  (41) 

and 

(42) 


I,) 

' 


(43) 


The  component  7^,  of  the  primary  current  is  called  the  mag- 
netizing current.  The  resultant  mutual  flux  <PMR  may  be  con- 
sidered due  to  7,,,  since  the  effect  of  the  other  component,  7/,  of 


MUTUAL-INDUCTION  189 

the  primary  current  is  balanced,  so  far  as  the  production  of  flux 
is  concerned,  by  the  secondary  current  J2.  Ii  and  J2  produce 
equal  and  opposite  ampere-turns,  equation  (41),  and,  therefore, 
cannot  cause  any  resultant  mutual  flux. 

In  an  air-core  transformer,  the  so-called  magnetizing  com- 
ponent, 7^,,  of  the  primary  current  is  very  large,  due  to  the  low 
permeability  of  the  magnetic  circuit.  If,  however,  the  frequency 
is  very  high,  the  flux  required  for  a  fixed  primary  voltage  will  be 
much  reduced.  Under  this  condition,  the  magnetizing  component 
of  the  primary  current  may  be  reduced  to  a  reasonable  value. 
For  an  iron-core  transformer  of  good  design,  the  component 
current  7^,  is  very  small  at  commercial  frequencies,  such  as  are 
used  for  power  generation  and  transmission.  If  the  primary 
resistance  and  primary  leakage  reactance  of  such  a  transformer 
are  made  small,  1^  will  be  nearly  constant  and  independent 
of  the  load,  varying  only  one  or  two  per  cent,  from  no  load  to 
full  load. 

Vector  Diagrams  of  an  Air-core  Transformer. — The  air-core 
transformer  is  an  important  example  of  two  inductively  coupled 
circuits  each  having  resistance  and  self-inductance.  The  sig- 
nificance of  the  components  in  the  equations  for  the  voltage 
induced  in  the  windings  of  circuits  having  resistance  and  self- 
and  mutual-inductance  will  be  made  clearer  by  a  study  of  the 
vector  diagram  of  an  air-core  transformer. 

Consider  an  air-core  transformer  with  a  load  on  its  secondary. 
The  impressed  voltage  drop  across  the  primary  will  be  equal  to 
the  primary  resistance  drop  plus  the  voltage  drops  in  the  primary 
winding  caused  by  its  self-inductance  and  by  the  mutual-induc- 
tance effect  of  the  secondary.  The  voltage  drop  across  the  sec- 
ondary will  be  equal  to  the  secondary  resistance  drop  plus  the 
voltage  drops  in  the  secondary  winding  caused  by  its  self-induc- 
tance and  the  mutual-inductance  effect  of  the  primary.  All 
equations  must  be  considered  in  a  vector  sense. 

The  voltage  drop  impressed  across  the  primary  winding  of  the 
transformer  is 

Vi  =     r,7i    +  jwLJi  +  jcoM/2  (44) 

EiR  =  juLj/i  +  juM  I  z  (45) 

-    .    ,  LJi       MI 

<PlR    =          <f>lL      + 


190 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Equation  (45)  is  an  equation  of  induced  voltage  drops.  Equa- 
tion (46)  is  an  equation  of  the  fluxes  producing  the  voltages  given 
in  equation  (45).  The  corresponding  voltages  and  fluxes  are 
placed  directly  under  each  other  in  these  two  equations. 

The  vector  diagram  corresponding  to  equation  (44)  is  shown 
in  Fig.  57.  The  primary  and  secondary  currents  are  assumed  to 
be  145  degrees  out  of  phase,  with  the  primary  current  leading. 


FIG.  57. 

The  numbers  of  turns  in  the  primary  and  secondary  windings 
are  taken  equal. 

If  the  primary  selfrinduced  voltage  drop,  JuLJi,  is  split  up 
into  two    parts,  one,  juSJi,  due   to   leakage   flux,  the  other, 

-  Nl 
rri  -fir,  due  to  the  component,  <PIM,  of  the  resultant  mutual 


flux  which  is  produced  by  the  primary  current,   equation   (44) 
becomes 


MUTUAL-INDUCTION  191 

V,  =  rJi  +  jo,  (SJi  +  M/ij^  +  j«J£L-          (47) 

The  total  induced  primary  voltage  drop,  l£i«,  i.e.,  the  voltage 
drop  produced  in  the  primary  winding  by  the  total  resultant 
primary  flux,  is  given  in  equation  (48).  Equation  (49)  shows  the 
component  fluxes  corresponding  to  the  component  voltage  drops 
given  in  equation  (48).  The  component  fluxes,  in  equation  (49), 
are  placed  directly  below  the  corresponding  component  voltage 
drops  in  equation  (48). 

~EiR  =  juSJi  +  juMIi        +  juMI*  (48) 


<PlR=          <PlS       +  (pl.M  +  V2M  (49) 

=       Vis  +  VMR  (50) 

The  total  primary  induced  voltage  drop,  EIR,  may  be  con- 
sidered to  be  made  up  of  two  components,  namely,  the  voltage 
drop  produced  by  the  primary  leakage  flux  and  a  voltage 
drop  produced  by  a  flux,  <PMR,  usually  called  the  mutual  flux, 
which  is  the  resultant  of  the  two  component  mutual  fluxes,  VIM 
and  VZM,  produced  by  the  primary  and  secondary  currents 
respectively.  The  voltage  corresponding  to  the  flux  <PMR  is 
marked  E±  on  Fig.  58.  This  voltage  is  usually  called  the  primary 
induced  voltage  drop.  In  reality,  it  is  only  a  part  of  the  primary 
induced  voltage  drop.  The  other  part  is  due  to  the  leakage 
flux  and  is  replaced  by  a  primary  leakage  reactance  drop,jxili  = 
jwSji  on  the  transformer  diagram  as  ordinarily  drawn.  The 
actual  primary  induced  voltage  drop  is  #IR.  Equations  exactly 
similar  to  equations  (48),  (49)  and  (50),  with  the  subscripts  1 
and  2  interchanged,  hold  for  the  secondary. 

Fig.  58  shows  the  component  fluxes  corresponding  to  the  terms 
of  equations  (48)  and  (49).  The  secondary  leakage  flux,  ^,  and 
the  flux,  VZL,  corresponding  to  the  self-induced  voltage  in  the 
secondary,  are  added.  VIR  =  VMR  +  <PIS  and  VZR  =  VMR  +  Vzs 
are  the  total  resultant  fluxes  linking  the  primary  and  secondary 
windings  respectively.  See  equations  (35)  and  (36),  page  186. 

As  was  stated  on  page  188,  the  primary  current  may  be  re- 
solved into  two  components:  one  I/,  which  is  opposite  to  the 
secondary  current  /2  and  just  balances  the  demagnetizing  effect 


192 


PRINCIPLES  OF  ALTERNATING!  CURRENTS 


of  that  current  in  producing  mutual  flux;  the  other  Iv,  which 
may  be  considered  to  produce  the  resultant  mutual  flux  ~VMR>  In 
an  air-core  transformer,  I,,  and  Ei  are  in  quadrature,  since  there 
is  no  core  loss.  The  only  losses  are  the  copper  losses.  These  are 
taken  care  of  toy  the  Jtfi  and  72r2  drops  in  the  primary  and 
secondary  windings.  In  Fig.  58,  //  =  —  72,  since  the  ratio  of 
the  turns  in  the  primary  and  secondary  windings  was  assumed  to 
be  unity,  i.e.,  Ni  was  assumed  to  be  equal  to  JV2-  Since  1^ 
produces  the  resultant  mutual  flux,  <PMR,  it  must  be  in  phase 
with  this  flux. 


FIG.  58. 


The  ordinary  vector  diagram  of  an  air-core  transformer  is 
shown  in  Fig.  59.  This  diagram  is  derived  directly  from  Fig.  58 
by  dividing  the  primary  current  into  the  two  components  // 
and  7^,  and  replacing  the  voltage  drops  due  to  the  leakage  in- 
ductances Si  and  >S2  by  the  leakage  reactance  drops  jliXi  and 
jHzXz.  Ei  is  the  voltage  drop  produced  in  the  primary  winding 
by  the  resultant  mutual  flux  VMR.  The  corresponding  voltage 
drop  in  the  secondary  winding  is  E2.  —Ez,  that  is  the  voltage 
rise,  is  used  on  Fig.  59  in  place  of  the  voltage  drop  to  keep  the 
diagram  more  open.  Also,  since  power  is  absorbed  on  the 
primary  side  of  a  transformer  and  delivered  on  the  secondary 
side,  it  is  better  to  draw  E\  as  a  voltage  drop  and  —Ez  as  a  volt- 


MUTUAL-INDUCTION 


193 


age  rise.  Since  E\  and  Ez  are  produced  by  the  same  flux,  i.e.,  by 
the  resultant  mutual  flux  <PMR,  they  must  be  in  phase  and  their 
magnitudes  must  be  in  the  same  ratio  as  the  number  of  primary 
and  secondary  turns.  E\  and  —Ez  are  therefore  in  opposite 
phase. 

EI      NI  /     . 

Wz   =   Wz  '       ' 

is  known  as  the  ratio  of  transformation  of  the  transformer. 


Load  Power  Factor  =  cos  62 
Primary  Power  Factor  =  cos  0, 


-Ea 


FlG.   59. 

In  Fig.  59,  Ei  is  equal  in  magnitude  to  —  Ez  since  Ni  and  Nz 
are  assumed  equal.  Since  power  is  delivered  on  the  secondary 
side,  drawing  —  Ez  as  a  voltage  rise  puts  its  energy  component  in 
phase  with  J2.  Drawing  Ei  as  a  voltage  drop  puts  its  energy 
component  in  phase  with  Ilm  The  secondary  terminal  voltage 
rise  is  equal  to  the  voltage  rise  —  Ez  minus  the  secondary  leakage 
reactance  and  resistance  drops. 

Example  of  the  Determination  of  the  Self-inductance,  Leak- 
age-inductance and  Mutual-inductance  of  an  Air -core  Trans- 
former.— When  the  secondary  winding  of  a  certain  air-core  trans- 
former is  on  open  circuit,  and  50  volts  at  60  cycles  are  impressed 
on  its  primary  winding,  it  takes  200  watts  at  0.7  power-factor. 
The  voltage  across  the  secondary  winding  under  these  conditions, 
measured  by  a  voltmeter  taking  negligible  current,  is  64.3  volts. 
The  primary  winding  has  2000  turns  and  the  ratio  of  the  number  of 

13 


194  PRINCIPLES  OF  ALTERNATING  CURRENTS 

turns  in  the  primary  and  secondary  windings  is  -vv-  =  0.5.     What 

are  the  self-inductance  and  leakage-inductance  of  the  primary 
winding  in  henrys?  What  is  the  mutual-inductance  of  the  two 
windings  in  henrys?  What  is  the  root-mean-square  or  effective 
voltage  induced  in  the  primary  winding  by  the  total  primary 
flux?  What  is  the  maximum  value  of  this  flux? 

VJi  cos  0i  =  PI 
50  X  /i  X  0.7  =  200 

200 


1  ~~ 


50  X0.7 
=  5.71  amperes 

Since  there  is  no  iron  core,  the  entire  power  taken  by  the  trans- 
former, when  there  is  no  load  on  the  secondary,  is  primary 
copper  loss. 

ri  =  j\  =  ^  71\2=  6-13  ohms- 

Taking  I\  as  the  axis  of  reals, 

Ji  =  5.71(1  +JO) 

Vi  =  50(cos  0i  +j  sin  0i) 

=  50(0.7  +  J0.7141) 

=  35  +  J35.71 

Since  there  is  no  power  consumed,  except  that  due  to  copper 
loss  in  the  primary  winding,  the  reactive  component  of  the  voltage 
drop  across  the  primary  winding  must  be  entirely  due  to  primary 
self-inductance. 

Ei  =  coLi/i  =  35.71  volts. 

or  71 


571 
Ei  =  «M/i 

fi4  ^ 

M  -  377-X  571 


MUTUAL-INDUCTION  195 

=  0.0166  -  0.0299  X  0.5 
=  0.0017  henry. 

E,  =  4.44/Vi^lO-8     (See  page  45.) 
#1 

<f>m    = 


35.71  X  108 

4.44  X  2000  X  60 

=  6.70  X  103  maxwells. 


CHAPTER  VII 

IMPEDANCES  IN  SERIES  AND  PARALLEL;  EFFECTIVE  RESISTANCE 
AND  REACTANCE 

Impedances  in  Series. — The  current  in  a  series  circuit  is 
the  same  in  all  parts  and  the  resultant  voltage  drop  across  the 
entire  circuit  is  equal  to  the  vector  sum  of  the  voltage  drops 
in  its  component  parts.  This  may  be  expressed  analytically  by 
the  following  equations. 

70  =  /1  =  /2  =  /,  =  etc.  (1) 

7o  =  #  i  +  E2  +  #3  +  etc.  (2) 

=  7ozi  +  7oz2  +  7oz3  +  etc.  (3) 

where  7o  and  Vo  are  the  resultant  current  and  the  resultant 
voltage  drop  across  the  entire  circuit.  7,  E  and  z  with  the 
subscripts  1,  2,  3,  etc.,  are  the  currents,  voltage  drops  and 
impedances  for  the  component  parts  of  the  circuit.  The  sum- 
mations in  equations  (2)  and  (3)  must  be  made  in  a  vector  sense. 

It  is  common  practice  to  use  a  dot  or  a  short  line  over  a  letter, 
when  the  latter  is  used  to  represent  a  vector  or  a  complex  quan- 
tity, in  all  cases  where  confusion  or  misunderstanding  would 
result  without  such  designation.  In  all  alternating-current  work 
currents  and  voltages  must  be  added  or  subtracted  vectorially. 
No  special  designation  therefore  is  necessary  in  most  cases  to 
indicate  this.  However,  for  the  sake  of  greater  clearness,  a  short 
line  will  be  used  over  all  letters  which  represent  either  vectors  or 
complex  quantities.  Power  is  not  a  vector,  neither  is  resistance 
nor  reactance,  but  resistance  and  reactance  drops  are  vectors. 
Impedance  is  a  complex  quantity.  Impedances,  therefore,  must 
be  handled  like  vectors  so  far  as  operations  of  addition,  subtrac- 
tion, multiplication  and  division  are  concerned. 

Consider  a  circuit  consisting  of  an  inductive  impedance,  ZL,  in 
series  with  a  capacitive  impedance,  zc.  The  diagram  of  connec- 
tions is  shown  in  Fig.  60a.  The  inductive  part  of  the  circuit  has  a 

196 


IMPEDANCES  IN  SERIES  AND  PARALLEL 


197 


resistance  rL  and  an  inductive  reactance  XL  =  2irfL  =  coL.     The 
capacitive  part  of  the  circuit  has  a  resistance  rc  and  a  capacitive 

reactance  xc  =  o~fC  =  ~~C'     The  vector  diagram  of  the  circuit  is 
shown  in  Fig.  606. 


(a) 


Referring  to  Fig.  606,  EL  =  IoVrL2  +  xL2  is  the  voltage  drop 
across  the  inductive  impedance  ZL.  This  drop  is  made  up  of  two 
parts:  one,  70rL  in  phase  with  the  current,  the  other,  IQXL  in 
quadrature  with  the  current  and  leading  it  by  90  degrees.  The 
voltage  drop  EL  leads  the  current  by  an  angle  whose  tangent  is 


~  =  ~-     EC  = 


^c2  -f-  xc*  is  the  voltage  drop  across  the 

capacitive  impedance  zc.     This  drop,  like  the  voltage  drop  EL 
across  the  inductive  impedance  ZL,  is  made  up  of  two  parts: 


one,  I0rc  in  phase  with  the  current,  the  other,  IQXC  =  —^  in 

quadrature  with  the  current  but  lagging  it  by  90  degrees.     The 
voltage  drop  Ec  lags  the  current  by  an  angle  Bc  whose  tangent  is 

x-c  _      -1 

rc        wrcC 

The  resultant  voltage  drop  Vo  across  the  two  impedances  in 
series  is  equal  to  the  vector  sum  of  the  voltage  drops  EL  and 


198  PRINCIPLES  OF  ALTERNATING  CURRENTS 

EC-  This  resultant  voltage  drop  is  made  up  of  two  parts,  /oro  = 
IO(TL  H-  rc)  and  70Zo  =  ?O(ZL  —  #c),  respectively  in  phase  and 
in  quadrature  with  the  current.  Whether  the  resultant  quad- 
rature component  leads  or  lags  the  current  depends  on  the  rela- 
tive magnitudes  of  the  inductive  and  capacitive  reactances,  XL 
and  —  xc.  If  XL  is  larger  than  —  xc  it  will  lead.  On  the  other 
hand,  if  —  xc  is  larger  than  XL  it  will  lag. 
From  Fig.  606  it  is  obvious  that 

EQ  COS  00    =    lofQ 

=  Io(rL  +  rc)  (4) 

EQ  sin  60    =  IQXQ 

=  Io(xL  -  XC)  (5) 

In  general,  if  there  are  k  impedances  in  series,  it  is  clear  from 
equations  (1)  and  (2)  that 

Eo  cos  80  =  I0r0  =  70(ri  +  r2  +    .    .    >+  rk)  (6) 

EQ  sin  0o  =  IOXQ  =  IQ(XI  +  xz  +   .    .    .  -f  xk)  (7) 
and 

r0  =  ri  +  r2  +     .    .    .    -f  r*  =  2*r  (8) 

TO  =  Xi  +  x2  +     .    .    .    +  xk  =  Z\x  (9) 

For  a  series  circuit,  therefore,  the  resultant  resistance  is  equal 
to  the  sum  of  the  resistances  of  its  separate  parts.  Similarly, 
the  resultant  reactance  of  a  series  circuit  is  equal  to  the 
sum  of  the  reactances  of  its  separate  parts.  A  series  circuit, 
therefore,  acts  like  a  simple  circuit  having  a  single  resistance  TQ 
in  series  with  a  single  reactance  XQ.  Resistances  are  always 
positive.  Reactances,  on  the  other  hand,  may  be  either  positive 
or  negative  according  as  they  are  due  to  inductance  or  to  capaci- 
tance. It  is  obvious,  therefore,  that  reactances  must  be  added 
algebraically,  i.e.,  with  regard  to  their  signs. 

Since  EQ  cos  00  =  /Oro  and  E0  sin  00  =  loXo  are  two  quadrature 
components  of  the  voltage  Eo,  it  follows  that 

EQ  =    7o\/ro2  +  XQZ  =  /0Zo  (10) 

and 


7     _ 


The  current  in  amperes  is  always  given  by  the  voltage  in  volts 
impressed  on  the  circuit  divided  by  the  resultant  impedance  in 


IMPEDANCES  IN  SERIES  AND  PARALLEL  199 

ohms.     The  resultant  impedance  z0  is  not  the  sum  of  the  sepa- 
rate impedances  except  in  complex. 


-  (xi  +  x2  -t-  .  .  .  -h  «§r 

=  V(SW2  +  (S*x)2  (12) 
Referring  to  Fig.  606,  it  will  be  seen  that 

tan    0o  =  -  (13) 

cos   00  =  ~°  (14) 

z0  /ICN 

sin    0o  =  —  {*•&) 

ZQ 

Complex  Method. — Since  the  current  is  the  same  in  all  parts 
of  a  series  circuit,  i.e.,  it  is  common  to  all  parts  of  the  circuit, 
it  will  be  taken  as  the  axis  of  reference,  that  is,  as  the  axis  of  reals, 

In  complex  the  voltage  drops  EL  and  Ec  across  the  two  parts 
of  the  series  circuit  shown  in  Fig.  60  are 

"EL  =  7oZL  =  7p(rL  +  jxL)  (16) 

>EC  =  IQZC  =  IO(TC  —  jxc)  (17) 

V0  =  EL  +  EC  =  [Q{  (TL  +  rc)  +  j(xL  -  xc) } 

=  7o(r0  +  jxo)  =  /o2o  (18) 

zo  =  r0  +  jx0  (19) 

XL   —    Xc          XQ  /nr\\ 

tan  00  =         i    r   =  ~  ^U^ 

ri  +  T4=  ,  =  T^  (21) 


V(rL  +  rc)«  +  (x,  -  *cY        * 

XL  -  XC  XQ 


,      . 


In  general,  if  there  are  k  impedances  in  series 
Ei  =  JoZi  =  Jo(ri  +  jxi) 
E2  =  lozz  =  7o(r2  +  3x2) 


Ek  =  loZk  =  Io(rk 


=  J0{  (ri  +  rz  +   .    .    .    +  rt) 

+  j(xi  +X2+   .    .    .    + 

(23) 


200  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  two  terms,  7o?*o  and  jloXo,  in  equation  (24),  are  respectively 
the  active  and  the  reactive  components  of  the  resultant  voltage 
drop  across  the  circuit 

tan  *„  =  Xf  =  |£  (25) 

TO  ^]/ 

cos  0o  =  -7^=  =  =  -°  =  -r-       2*r  (26) 

Vro2  +  zo2       zo        V  (Z*r)2  + 


sin  00  =  —j=  =  =        =  -      ^-  (27) 

Vr02  +  *o2       zo       V  (2W2  +  (S?z)2 

Although  the  resistances  and  the  reactances  of  a  series  circuit 
may  be  added  directly  to  give  respectively  the  resultant  re- 
sistance and  reactance,  impedances  may  not  be  so  added. 
Impedances  must  always  be  added  in  complex.  For  example, 

Z0   =  Zi   +  Z2  +     .      .      .      +  Zjfc 

can  only  mean 

20  =  (ri  +  r2  +    .    .    .    +  rk)  +j(Xl  +  z2  +    .    .    .   +  xk)     (28) 

It  can  never  mean  anything  else.     The  resultant  impedance  in 
ohms  is 

(29) 


+  z02  (30) 

where  the  r's  and  x's  are  expressed  in  ohms. 

When  only  the  ampere  value  of  the  current  in  a  series  circuit 
is  desired,  it  is  found  by  dividing  the  magnitude  of  the  impressed 
voltage  in  volts  by  the  magnitude  of  the  resultant  impedance  in 
ohms. 

EQ  (volts)        T    , 

-^—r   —  4  =  /0  (amperes).  (31) 

z0  (ohms) 

The  phase  of  the  current  with  respect  to  the  voltage  is  fixed 
by  equations  (25),  (26)  and  (27),  in  which  00  is  the  angle  of 
lead  of  the  voltage  with  respect  to  the  current.  The  angle  of  lead 
of  the  current  with  respect  to  the  voltage  would  be  —  do.  Since 
a  negative  angle  of  lead  is  equivalent  to  an  angle  of  lag,  the  angle 
Bo  in  equations  (25),  (26)  and  (27)  may  equally  well  be  considered 
as  the  angle  of  lag  of  the  current  with  respect  to  the  voltage. 

Whether  00  actually  represents  an  angle  of  lead  or  an  angle  of  lag 
in  any  particular  case  will  depend  upon  its  sign  as  fixed  by  the 


IMPEDANCES  IN  SERIES  AND  PARALLEL  201 

resultant  reactance  of  the  circuit.  When  00,  as  determined  by 
the  resultant  reactance  XQ,  is  positive,  i.e.,  when  the  sum  of  the 
inductive  reactances,  2xL  =  22-jrfL,  is  greater  than  the  sum  of  the 

capacitive  reactances,  S  (—  xc)  =  2'  ^°  *S  actuan<y  an  angle  of 


lead  of  the  voltage  with  respect  to  the  current  or  an  angle  of  lag 
of  the  current  with  respect  to  the  voltage.  In  this  case  the  volt- 
age leads  the  current  or  the  current  lags  the  voltage.  The  circuit 
as  a  whole  is  inductive.  When  00  is  negative,  it  is  actually  an 
angle  of  lag  of  the  voltage  with  respect  to  the  current  or  an  angle 
of  lead  of  the  current  with  respect  to  the  voltage.  In  this  case  the 
voltage  lags  the  current  or  the  current  leads  the  voltage.  The 
circuit  as  a  whole  is  capacitive. 

Example  of  Impedances  in  Series.  —  A  circuit,  which  consists 
of  two  impedances  and  a  non-inductive  resistance  in  series,  is 
connected  across  a  230-volt,  60-cycle  circuit.  One  of  the  im- 
pedances has  a  self  -inductance  LI  =  0.1  henry  and  a  resistance 
7*1  =  5  ohms.  The  other  impedance  has  a  capacitance  C  =  100 
microfarads  and  a  negligible  resistance.  The  non-inductive 
resistance  r&  =  10  ohms.  What  is  the  current  and  what  is  its 
phase  with  respect  to  the  impressed  voltage?  How  much  power 
does  the  circuit  absorb  and  what  is  the  power-factor?  What  are 
the  potential  drops  across  each  of  the  impedances  and  the  non- 
inductive  resistance,  and  what  are  the  phase  angles  of  these 
drops  with  respect  to  the  current  in  the  circuit?  What  are  their 
phase  angles  with  respect  to  the  impressed  voltage? 

7*1  =  5  ohms 

xi  =  27r60  X  0.1  =  377  X  0.1  =  37.7  ohms. 

r2  =  0 

—  106 


r3  =  10  ohms. 

xz  =  0 

zo  =  (n  +  r2  +  r3)  +  j(xi  +  xz  +  x3) 

=  (5  +  0  +  10)  +  j(37.7  -  26.53  +  0) 

=  15+J11.17 

zo  =  V(15)2  +  (11.  17)2 

=  18.70  ohms. 

230 
/o  =      -      =  12.30  amperes. 


202  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Power-factor  =  cos  00  =  -  =  -^|n  =  0.8021 

ZQ         1O.7U 

0o  =  36.67  degrees. 

Total  power  =  P  =  230  X  12.30  X  0.8021 
=  2264  watts. 

The  voltage  impressed  on  the  circuit  leads  the  current  by 
36.67  degrees  or  the  current  lags  the  voltage  by  the  same  angle. 
The  circuit  as  a  whole  is  inductive,  since  XQ  is  positive. 


EI  (drop  across  Zj)  = 

=  12.30  X  V(5)2  +  (37.7)2 
=  12.30  X  38.03 

=  467.8  volts. 
07  7 

tan  0!  =  ^r-  =  7.54 
o 

0i  =  82.45  degrees. 

The  voltage  drop  Ei,  across  the  inductive  impedance,  leads 
the  current  by  82.45  degrees. 

Ez  (drop  across  zz)  =  IoZ2 

=  12.30  X  A/(0)2  +  (26.53)2 

=  12.30  X  26.53 

=  326.3  volts. 

-26.53 

tan  02  =  —  ~  --  =  —  oo 

02  =  -90  degrees. 

The  voltage  drop  Ez,  across  the  capacitive  impedance,  lags  the 
current  by  90  degrees. 


(drop  across  23)  = 

=  12.30  X  V(lW2~+W2 
=  12.30  X  10 
=  123.0  volts, 

tan  03  =  j  Q  =  0 
03  =  0  degrees, 


IMPEDANCES  IN  SERIES  AND  PARALLEL  203 

The  voltage  drop  E?.,  across  the  non-inductive  resistance,  is  in 
phase  with  the  current. 

Since  the  voltage  impressed  across  the  circuit  leads  the  current 
by  36.67  degrees,  to  get  the  phase  of  the  voltage  drops  across 
the  impedances  and  the  non-inductive  resistance  with  respect 
to  the  impressed  voltage  Fo,  36.67  degrees  must  be  subtracted 
from  the  phase  angles  of  the  drops  with  respect  to  the  current. 

Therefore,  the  phase  angles  of  the  drops  with  respect  to  the 
voltage  are 

0i'  =  82.45  -  36.67  =  45.78  degrees. 
02'  =  -90.0  -  36.67  =  -126.7  degrees. 
03'  =    o.O     -  36.67  =  -  36.67  degrees. 

Ei  leads  the  voltage  drop  Fo  by  45.78  degrees. 

E%  leads  the  voltage  drop  Fo  by  —126.7  degrees  or  lags  it 
by_126.7  degrees. 

Es  leads  the  voltage  drop  Fo  by  —36.67  degrees  or  lags  it  by 
36.67  degrees. 

It  should  be  noted  that  the  voltage  drop  across  each  impe- 
dance is  much  greater  than  the  voltage  drop  across  the  entire 
circuit.  This  is  always  possible  when  a  circuit  contains  both 
inductance  and  capacitance  in  series  with  a  low  resistance.  In 
this  particular  problem  the  drops  across  the  two  impedances  are 
82.45  +  90.0  =  172.45  degrees  apart  in  phase  and  therefore 
contribute  little  to  the  resultant  voltage  across  the  circuit. 
More  will  be  said  about  this  condition  under  "  Resonance." 

The  preceding  problem  could  have  been  solved  by  the  complex 
method,  but  in  this  particular  problem  the  ordinary  algebraic 
method  gives  a  somewhat  shorter  solution. 

Another  Example  of  Impedances  in  Series. — A  certain  circuit 
consisting  of  two  impedances,  one  of  which  has  a  resistance  of  10 
ohms  and  an  inductive  reactance  of  10  ohms,  takes  a  leading 
current  of  15  amperes  at  0.8  power-factor  when  connected 
across  a  200- volt,  60-cycle  circuit.  What  are  the  resistance  and 
reactance  of  the  second  impedance? 

The  use  of  the  complex  method  will  give  the  most  direct  solu- 
tion in  this  case.  Call  the  known  impedance  Zi  and  the  unknown 
impedance  z2.  Take  the  current  as  the-  axis  of  reference.  Then 


204  PRINCIPLES  OF  ALTERNATING  CURRENTS 

To    =  15(1  +  JO)  =  15+ JO 
Fo  =  200(cos  0o  -  j  sin  00) 

=  200(0.8  -  jO.6) 

=  160  -  J120 

V0      160  -  J120       - 


Jo 

=  10.67  -  J8.0- 
r0  =  10.67  ohms. 
XQ  =  —8.0  ohms. 
r2  =  10.67  -  n 

=  10.67  -  10.00  =  0.67  ohm. 
Xz  =  — 8.0  —  x\ 

=  -8.0  -  10.0 

=  -18.0  ohms. 

The   second  impedance   is  therefore   capacitive,   since  Xz  is 
negative.     Its  complex  expression  is 

z2  =  0.67  -  J18.0 

It  has  a  resistance  of  0.67  ohm  and  a  capacitive  reactance  of 
18.0  ohms.     If  Xz  is  expressed  in  ohms  and  Cz  in  microfarads 

-106 

*2  27T/C2 

-106 


27r60(-18.0) 
=  147.4  microfarads. 

Series  Resonance. — A  series  circuit  containing  inductance 
and  capacitance  is  said  to  be  in  resonance  when  the  sum  of  the 
inductive  reactances  is  equal  to  the  sum  of  the  capacitive  reac- 
tances. For  resonance  therefore  . 

X0  =  zi  +  x2  +  z3  +  .    .    .    +  xk  =  0  (32) 

=  ZxL  -  2xc  =  0  (33) 

Consider  a  circuit  containing  a  resistance  r,  an  inductance 
L  and  a  capacitance  C,  in  series. 

E 


I  = 


IMPEDANCES  IN  SERIES  AND  PARALLEL  205 

For  resonance, 

<oL  =  4>or  co2LC  =  1  (34) 

coC 

and 

(35) 


With  constant  frequency,  resonance  may  be  produced  by 
adjusting  either  L  or  C  or  both,  equation  (34),  or  with  L  and  C 
constant,  it  may  be  produced  by  adjusting  the  frequency,  equa- 
tion (35).  Since  the  resultant  reactance  is  zero,  the  current  at 
resonance  is  in  phase  with  the  voltage  impressed  across  the  entire 
circuit  and  follows  Ohm's  law.  It  is  given  by 

'    H       '-7 

Since  the  resultant  reactance  at  resonance  is  zero,  the  resultant 
impedance  is  a  minimum  and  is  equal  to  the  resultant  resistance 
of  the  circuit.  The  current  is  therefore  a  maximum  and 
is  limited  only  by  the  resultant  resistance  of  the  circuit.  It  is 
entirely  independent  of  the  magnitudes  of  the  inductive  and 
capacitive  reactances. 

The  drop  in  potential  due  to  the  inductance  is  7coL.     Due  to  the 

capacitance  it  is  —  •     These  two  potential  drops  at  resonance  are 

without  effect  on  the  resultant  potential  drop  across  the  circuit  as 
a  whole.  They  may  be  much  greater  than  the  voltage  impressed 
on  the  entire  circuit  and  may  easily  reach  excessive  values  when 
the  resultant  resistance  of  the  circuit  is  small  in  comparison  with 
its  inductive  and  capacitive  reactances.  Series  resonance,  for 
constant  potential  circuits,  is  usually  a  very  undesirable  condi- 
tion, on  account  of  the  excessive  current  and  high  voltages  that 
may  be  produced  when  the  resistance  is  small  compared  with  the 
inductive  and  capacitive  reactances. 

If  a  low  resistance  series  circuit  is  tuned  for  resonance  by 
varying  the  capacitance,  and  curves  of  current  and  of  voltage 
drops  across  the  inductance  and  capacitance  are  plotted  against 
capacitance,  they  will  all  show  very  marked  peaks  at  resonance. 
The  maximum  current  and  maximum  voltage  drop  across  the 
inductance  will  occur  at  resonance,  but  the  maximum  voltage 


206 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


drop  across  the  capacitance  will  not  occur  at  resonance.  If  the 
capacitance  is  constant  and  the  inductance  is  varied,  the  maxi- 
mum drop  across  the  capacitance  will  then  occur  at  resonance,  but 
now  the  maximum  voltage  drop  across  the  inductance  will  not 
occur  at  resonance.  The  steepness  with  which  the  curves  rise  as 
resonance  is  approached  depends  upon  the  magnitude  of  the 
resistance  of  the  circuit  compared  with  the  magnitudes  of  the 
inductive  and  capacitive  reactances.  The  point  of  resonance  is 
most  pronounced  when  the  resistance  is  low  in  comparison  with 
the  reactances. 


V=  100  volts 
/  =  60  cycles 
L  =  0.1  henry 
r  =  1,  4  and 

8  ohms  for 

curves  tt 

b  and  c 

respectively 


50        60        70        80         90        100      110       120 
Capacitance  in  Microfarads 

FIG.  61. 

When  a  series  circuit  is  tuned  for  resonance  by  varying  either 
L,  C  or  /,  low  resistance  permits  very  sharp  tuning.  With  high 
resistance,  the  resonance  curves  become  flattened  and  sharp 
tuning  is  then  impossible.  With  high  resistance  compared  with 
the  reactances,  the  point  of  resonance  will  not  be  well  defined. 
This  is  illustrated  in  Fig.  61. 

Three  current  curves,  showing  series  resonance,  are  plotted  in 
Fig.  61  for  a  60-cycle,  100-volt  circuit  having  an  inductance  of 
0.1  henry  and  a  variable  capacitance.  Curve  a  is  for  a  resistance 
of  1  ohm.  Curves  b  and  c  are  for  resistances  of  4  and  8  ohms 
respectively.  The  lower  parts  of  the  three  curves  so  nearly 


IMPEDANCES  IN  SERIES  AND  PARALLEL 


207 


coincide,  that  for  b  and  c  only  the  portions  of  the  curves  near 
resonance  are  plotted.  For  curve  a  the  voltage  drop  across  "the 
inductance  is  3770  volts  at  resonance.  It  is  one-quarter  and 
one-eighth  this  amount  at  resonance  for  curves  b  and  c  respec- 
tively. 


1800 
1600 
1400 

Juoo 

8*1000 

c 
5 
a  BOO 


400 

200 

0 

-200 
-400 
-600 

!- 

l-iooo 

g-1200 
-1400 
-1600 
-1800 


0.4      0.6      0.8      1.0      1.2      14^^    1.8      2.0     2.2     24 
Frequency,  Cycles !  per  sec.  xjlO'6 


Voltage         =  1  volt 
Capacitance  -  0.00235  microfarad 
Inductance    =  377  microhenrya 
Resistance     »  25  ohms 


30 


40 


10 


FIG.  62. 


Fig.  62  shows  the  effect  of  varying  the  frequency  to  produce 
resonance  in  a  circuit  containing  constant  resistance,  constant 
inductance  and  constant  capacitance  in  series.  Curves  of 


inductive  reactance,  XL  =  uL,  capacitive  reactance,  xc  = 


208  PRINCIPLES  OF  ALTERNATING  CURRENTS 

resultant    reactance,    XL  —  xc  =  uL  —  — >    and    current    are 

Ceo 

plotted  for  a  circuit  having  r  =  25  ohms,  L  =  377  microhenrys 
and  C  =  0.00235  microfarads.  Resonance  occurs  at  a  fre- 
quency of  1.69  X  105  cycles. 

Free  Period  of  Oscillation  of  a  Resonant  Circuit  Containing 
Constant  Resistance,  Constant  Inductance  and  Constant  Ca- 
pacitance in  Series.— The  natural  or  free  oscillation  frequency 
of  a  circuit  containing  constant  resistance,  constant  inductance 
and  constant  capacitance  in  series  is 

,  =  _, JL  (36) 


(See  equation  124,  page  159.)     When  r2  is  small  compared  with 

4L 

—  equation  (36)  reduces  to 

C 


Therefore,  the  resonant  frequency  of  a  series  circuit,  which 
has  low  resistance  compared  with  the  ratio  of  its  inductance  and 
capacitance,  is  the  same  as  its  free  oscillation  frequency.  This 
makes  it  possible  to  tune  a  series  circuit  containing  inductance 
and  capacitance  to  have  a  sharp  current  maximum  for  a  definite 
frequency. 

A  circuit  on  which  a  non-sinusoidal  electromotive  force  is- 
impressed  may  be  in  resonance  for  a  certain  harmonic  and  yet 
be  very  far  from  resonance  for  the  fundamental  or  other  har- 
monics (see  equation  (35),  page  205).  A  circuit  can  be  in 
resonance  for  only  one  frequency  at  the  same  time.  By  proper 
tuning,  as  by  varying  either  the  inductance  or  the  capacitance 
or  both,  it  is  possible  to  exaggerate  any  harmonic  in  the  current 
for  which  there  is  a  corresponding  harmonic  in  the  impressed 
voltage.  Tuning  is  made  use  of  in  radio-telegraphy  in  order  to 
make  a  receiving  circuit  respond  to  electromagnetic  waves  of  a 
definite  frequency.  On  account  of  the  very  high  frequency  used 
in  radio-communication,  the  resistance  of  radio  circuits  is  usually 
very  small  in  comparison  with  their  reactances,  making  very 
sharp  tuning  possible, 


IMPEDANCES  IN  SERIES  AND  PARALLEL  209 

A  Problem  in  Resonance. — A  series  circuit,  which  consists  of  a 
non-inductive  resistance  of  5  ohms,  an  impedance  of  1.5  ohms 
resistance  and  0. 1  henry  inductance  and  a  variable  capacitance, 
is  connected  across  a  constant  potential  220-volt,  60-cycle  cir- 
cuit. If  the  capacitance,  starting  with  a  value  too  small  to 
produce  resonance,  is  gradually  increased,  for  what  value  of  the 
capacitance  will  the  voltage  drop  across  it  have  the  greatest 
value?  For  what  value  of  the  capacitance  will  the  voltage  drop 
across  the  inductance  have  the  greatest  value?  As  the  capacit- 
ance is  increased,  which  of  the  two  voltage  drops  will  have  its 
maximum  value  first? 

XL  =  377  X  0.1  =  37.7  ohms 


VL  =  IwL  =  IxL 

where  Vc  and  VL  are  the  potential  drops  across  the  capacitance 
and  inductance  respectively. 


Vro2  +  (XL  -  Xc)2 
Vc      Ixc        /               V°              -t 

°'        lVr.'+(«t 

-  *c)2J 

VrV+  (XL  -  xcY}X 
(XL  ~  XC)2 


dxc  ro2  +  (XL  -  xc)2 

=  0 

r02  +  (XL  -  xc)2  +  XC(XL  -  xc)  =  0 
r02  +  xL2  —  2xLxc  +  xc2  +  XLXC  —  xc2  =  0 


Xc  = 


ro2  +  xL2 


+  37.7  =  38.8  ohms. 


106  10 


377  X  38.8 
=  68.4  microfarads. 

The    capacitance    for    maximum    voltage    drop    across    the 
capacitance  is  therefore  68.4  microfarads. 

14 


210  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  maximum  voltage  drop  across  the  capacitance  is 

220 

Vc  (maximum)  =       ,  =  =   X  38.8 

V(6.5)2  +  (37.7  -  38.8)2 

220    X  38.8 


6.593 
=  1293  volts. 

Since  L  is  constant,  the  voltage  drop  across  the  inductance 
must  be  a  maximum  when  the  current  is  a  maximum.  The 
current  is  a  maximum  at  resonance.  Therefore 

•\7 

VL  (maximum)  =  —  -  XL 
TO 

-22°X377 
6.5  X 

=  1276  volts. 

At  resonance  Vc  and  VL  are  equal  in  magnitude.  Vc  at 
resonance  is  therefore  1276  volts,  which  is  less  than  its  maximum 
value. 

—  XC  =  XL 

C  (at  resonance)  = 


=  70.3  microfarads. 

The  voltage  drop  across  the  capacitance  reaches  its  maximum 
before  the  voltage  drop  across  the  inductance  reaches  its  maxi- 
mum. As  resonance  is  approached  (C  increasing,  xc  decreasing) 
the  current  rises  very  rapidly  for  a  small  decrease  in  xc.  The 
increase  in  the  current  will  more  than  balance  the  decrease  in  xc 
until  the  point  of  resonance  is  nearly  reached.  As  resonance  is 
approached,  the  current  curve  flattens  out  preparatory  to  de- 
creasing after  resonance  is  passed.  At  some  point  just  before 
resonance  is  reached,  the  decrease  in  xc  will  just  balance  the  in- 
crease in  7.  The  maximum  voltage  drop  across  the  capacitance 
will  occur  at  this  point.  When  the  resistance  is  small  in  com- 
parison with  XL,  the  maximum  voltage  drop  across  the  capacitance 
will  occur  very  near  the  point  of  resonance.  When  the  resistance 
is  large,  the  maximum  voltage  drop  across  the  capacitance  may 
occur  quite  a  bit  before  resonance  is  reached. 

In  the  problem  just  solved,  what  additional  resistance  will  it 
be  necessary  to  place  in  series  with  the  circuit  in  order  that  the 
maximum  voltage  drop  across  the  capacitance  shall  be  limited  to 
500  volts. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  211 

The  capacitive  reactance  for  maximum  voltage  drop  across  the 
capacitance  is  (equation  (38)) 

ro2  +  x# 

Xc"     ~ 


The  resultant  reactance  is 

XQ  =  XL-  Xc  -.=  XL  - 


ro2  +  XL*  r02 


The  resultant  impedance  is 


The  resultant  impedance  drop  across  the  condenser  is 

V c  =  Ixc  =  -  -  X  xc 

ZQ 

220         ,  r02  -f  xL2      _nn      ., 
= X =  500  volts 


Vr02  +  xL2  =  500 
r0  220 

XL  37.7 


/500\ 2  / /500\ 2 

\220/  \  \220/ 

=  18.5  ohms  total  resistance. 
Added  resistance 

=  18.5  -  1.5  -  5.0 
=  12.0  ohms. 

Impedances  in  Parallel. — The  voltages  impressed  across  the 
branches  of  a  circuit  consisting  of  a  number  of  impedances  in 
parallel  are  equal.  The  resultant  current  taken  by  the  circuit 
is  equal  to  the  vector  sum  of  the  component  currents  taken  by  the 
branches.  This  may  be  expressed  analytically  as  follows: 

y0  =  EI  =  E2  =  E*  =  etc.  (39) 

=  IiZi  —  IzZz  =  /s2s  =  etc.  (40) 

7o  =7i  +  /2  +  73_+etc.  (41) 

'   0  '0  '    0  /  j  f\\ 

21  22  23 

fi  +  i+  i  +  etc.}  (43) 

L3l  22  23 


212 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Consider  a  circuit  consisting  of  two  impedances,  z\  and  22, 
in  parallel  as  indicated  in  Fig.  63a. 

The  impedance,  z\,  of  branch  1  has  a  resistance  ri  and  an 
inductive  reactance  Xi  =  coZ/i.  The  impedance,  22,  of  branch  2 


-1 


has  a  resistance  r2  and  a  capacitive  reactance  x2  =  —g  '    Since 

the  impedances  are  in  parallel  the  same  potential,  F0,  is  impressed 
across  their  terminals. 


0       ?! 


(a) 


FIG.  63. 


The  graphical  construction  for  this  circuit  is  shown  in  Fig. 
636.  Fo  is  the  voltage  drop  impressed  on  the  circuit  and  is  also 
the  potential  drop  across  each  of  the  branches.  (Equation  (39).) 
Since  these  drops  are  equal  and  in  phase,  F0  must  be  the  common 
side  of  the  right-angled  voltage  triangles.  The  intersection  of 
the  vectors  representing  the  resistance  and  reactance  drops  in 
each  of  the  branches  of  the  circuit  must  therefore  lie  on  a  circle 
drawn  with  the  vector  Fo  as  diameter. 

All  voltages  are  drawn  to  a  common  scale.  Likewise  all 
currents  are  drawn  to  a  common  scale,  but  the  scales  for  voltage 
and  current  may  be,  and  indeed  usually  will  be,  different. 

The  sides  OD  and  DF0  of  the  resultant  triangle  ODVQ  are  the 
resultant  resistance^  and  reactance  drops,  r07o  and  z07o,  respec- 
tively. Knowing  70,  the  resultant  resistance,  r0,  and  the  re- 
sultant reactance,  XQ,  may  be  found. 

From  the  diagram  (Fig.  636) 


/o  cos  00  =  Ii  cos  0i  +  Ii  cos  02 
/o  sin  00  =  /i  sin  0]  +  72  sin  02 


(44) 
(45) 


IMPEDANCES  IN  SERIES  AND  PARALLEL  213 


These  equations  may  be  written 


0    V                           °\Xl_l_         *  \S   '* 

~   /\                         /\                   ~  A.  ~ 
2o           ZQ           Z\           Z\          Z%            Z% 

V    V         r°              V    1        Tl         4-         Tz          1 

(46) 

(47) 
(48) 
(49) 

(50) 

CK11 

'OA          0,             „             >l)j         21^.2'            21             2/ 

TQ    -p  XQ                1  7"i    -f-  Xi          r2    -f-  X2     I 

2g           2o           2i            Zi           Zi            22 
T7      X             X°               -    V      1            Xl               I               ^             1 

^OA2.2              '  •   1         21             2'            21              2l 

TO   'h  ZQ               I**   T  Zi        r2   ~r  #2    J 
Hence 

r02  +  z02      ri2  +  ii2   '   r22  +  x22 

In  general,  if  there  are  k  branches  in  parallel 


P--^FTP  (53) 

It  must  be  remembered  that  x  is  positive  for  inductance  and 
negative  for  capacitance.  Therefore  the  summations  in  equa- 
tions (45),  (48),  (49),  (51)  and  (53)  must  be  made  in  an  algebraic 
sense. 

Conductance,  Susceptance  and  Admittance  of  a  Circuit.  —  The 
expression 

r-4^  =  9         '•     '  '  (54) 

is  known  as  the  conductance  of  a  circuit  and  is  denoted  by.  the 
letter  g.  When  there  are  a  number  of  circuits  in  parallel,  their 
resultant  conductance,  from  equation  (52),  is  given  by  the  sum  of 
their  separate  conductances.  Therefore 

go  =  0i  +  02  +   .    .    .    +  Qk 

=  X*g  (55) 

The  expression 

-  6  (56) 


214  PRINCIPLES  OF  ALTERNATING  CURRENTS 

is  known  as  the  susceptance  of  a  circuit  and  is  denoted  by  the 
letter  6.  When  there  are  a  number  of  circuits  in  parallel,  their 
resultant  susceptance,  from  equation  (53),  is  given  by  the  sum  of 
their  separate  susceptances.  Since  reactance,  x,  may  be  either 
positive  or  negative,  according  to  whether  it  is  inductive  or 
capacitive  reactance,  it  is  obvious  from  the  expression  for  sus- 
ceptance, equation  (56),  that  susceptance  is  likewise  either 
positive  or  negative.  Inductive  susceptance  is  positive.  Ca- 
pacitive susceptance  is  negative.  Since  susceptance  may  be 
either  positive  or  negative,  susceptances  must  always  be  added 
in  an  algebraic  sense,  i.e.,  with  regard  to  their  signs. 

bo  =  bi  +  &2  +    .    •    •    +  bk 

=  ^b  (57) 

Since  70  cos  00  =  V0go  and  70  sin  00  =  Vob0  are  two  quadrature 
components  of  the  resultant  current  70  (see  Fig.  636),  it  is  obvious 
that 


/o  =  V0     </o2  +  &o2  (58) 

o  and  VobQ  are  the  active  and  reactive  components  of  the 
current  /o  with  respect  to  the  voltage  V0. 
The  expression 


2/o  =  v        +  &o2  (59) 

is  known  as  the  admittance  of  the  circuit  and  is  denoted  by  the 
letter  y.  From  equations  (58)  and  (59) 

2/o  =  ~  (60) 

Impedance  is  ZQ  =  -7^.     Admittance,  therefore,  is  the  reciprocal 

-to 

of  impedance.  Although  admittance  is  always  the  reciprocal  of 
impedance,  conductance  is  never  the  reciprocal  of  resistance 
except  when  the  reactance  is  zero.  (Equation  (54).)  Suscep- 
tance is  never  the  reciprocal  of  reactance  except  when  the  resistance 
is  zero.  (Equation  (56).) 

Admittance  and  also  conductance  and  susceptance  are 
measured  in  reciprocal  ohms.  This  unit  is  called  the  mho,  the 
word  ohm  written  backwards.  When  resistance  and  reactance 
in  equations  (54)  and  (56)  are  expressed  in  ohms,  the  conductance 
and  susceptance  will  be  in  mhos.  When  /0  and  Vo  in  equation 
(60)  are  in  amperes  and  volts  the  admittance  2/0  is  in  mhos. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  215 

The  current  in  a  series  circuit  in  amperes  is  given  by  the  resul- 
tant impressed  voltage  in  volts  divided  by  the  resultant  impedance 
of  the  circuit  in  ohms.  The  resultant  current  in  a  parallel 
circuit  in  amperes  is  given  by  the  impressed  voltage  in  volts 
multiplied  by  the  resultant  admittance  of  the  circuit  in  mhos. 
For  the  parallel  circuit 

tan  0o  =  — >      cos  00  =  —  >      sin  00  =  — •  (61) 

go  .  2/o  2/o 

Vector  Method. — From  equation  (43),  page  211 


fl     1     1 

Fo  I  7  +  ~  4-  r  +  etc. 


7o  =  Fo  \  ~  4-  -  4-  ~  +  etc.  (62) 

[  Z\         Z2         Zz 

Putting  the  z's  in  their  complex  form,  equation  (62)  becomes 

/0  =   yl--L+_l_+__l         +et(j  1  (63) 

Ifj  4-  JXi       r2  +  jx2      r3  +  jxs 

Rationalizing, 


it  =  Fo  _  x 

\r\  4-  jxi 


r3  +  jx3       r3  - 

4- 


X!  r 

.      _  +(r-7f^-J'77f^')  +  etc-!      (64) 

==  Vo  {  (gi  +  02  +  ft  +  etc.)  -  j(6x  +  62  +  63  4-  etc.)  |  (65) 

=  Fo  (0o  -  jbo) 

go  =  0i  +  02  4-  #3  4-  etc.  (66) 

60  =  61  4-  b2  4-  63  4-  etc.  (67) 

It  should  be  noted  that  while  impedance  is  z  =  r  4-  jz,  admit- 
tance is  y  =  g  —  jb.  Since  inductive  reactance  is  positive  and 
capacitive  reactance  is  negative,  the  impedance  of  an  inductive 
circuit  is  actually  z  =  r  -}-  jx  and  the  impedance  of  a  capacitive 
circuit  is  actually  z  =  r  —  jx.  Since  inductive  susceptance  is 
positive  and  capacitive  susceptance  is  negative,  the  admittance 
of  an  inductive  circuit  is  actually  y  =  g  —  jb  and  the  admittance 


216  PRINCIPLES  OF  ALTERNATING  CURRENTS 

of  a  capacitive  circuit  is  actually  y  =  g  +  jb.  The  signs  of  the 
second  terms  in  the  complex  expressions  for  impedance  and 
admittance  are  opposite. 

For  impedances  in  parallel,  the  resultant  conductance  is  the 
sum  of  the  component  conductances  and  the  resultant  suscep- 
tance  is  the  algebraic  sum  of  the  component  susceptances.  Con- 
ductances, susceptances  and  admittances,  as  has  already  been 
stated,  are  measured  in  reciprocal  ohms,  i.e.,  in  mhos. 

Admittances  cannot  be  added  directly.  Admittance  is  a 
complex  quantity.  Admittances,  therefore,  like  impedances, 
can  only  be  added  in  complex.  For  example: 

2/0  =    y\    +  2/2    +   2/3    +      .       .       •      +  2/Jfc 

can  only  mean 

2/0  =  2/1  +  2/2  H-  2/3  +  .   .   .  +  27* 

=    (01    +  #2    +  03    +      .       -       -      +   fiffc) 

-j(bi  +  62  +  63  +   .    .    .    +6*)  (68) 

It  can  never  mean  anything  else. 

Resistance  and  Reactance  in  Terms  of  Conductance  and 
Susceptance. — It  is  often  necessary,  when  dealing  with  a  parallel 
circuit,  to  determine  its  resultant  resistance  and  reactance,  i.e., 
the  resistance  and  reactance  of  the  simple  series  circuit  which 
would  replace  it  so  far  as  total  power,  resultant  current  and  power- 
factor  are  concerned.  The  expressions  for  r  and  x  in  terms  of  g 
and  b  may  be  found  from  the  relation  between  impedance  and 
admittance 

z  =  r  +  jx 


y     g  - 
1         g  +  ft 


-jrJFp  +  JjTfp  (68) 

Evidently 


IMPEDANCES  IN  SERIES  AND  PARALLEL  217 

Polar  Expression  for  Admittance. — It  is  often  convenient, 
when  dealing  with  the  more  complicated  problems  such  as 
those  associated  with  the  transmission  line,  to  use  the  polar 
expression  for  admittance.  The  polar  expression  for  impedance 
is  given  in  Chapter  V,  pages  136,  150  and  170. 


J-'-'  +  6lj 

(72) 
(73) 

The  expression  y\~~0  or  y\B  is  known  as  the  polar  expression 
for  admittance.  The  angle  0  is  the  angle  of  the  admittance  and 
is  determined  by  the  relation 

0  =  tan-1  -  (74) 

g 

Admittance  is  a  scalar  quantity,  y,  multiplied  by  an  operator 
0  =  (cos  0  —  j  sin  0),  which  rotates  through  an  angle  —0. 

When  vector  voltage  is  multiplied  by  admittance  either  in 
complex  or  polar  form,  it  gives  vector  current  in  its  proper  phase 
relation  with  respect  to  the  voltage,  regardless  of  the  axis  to 
which  the  voltage  is  referred. 

An  Example  of  the  Solution  of  a  Parallel  Circuit  by  the  Use 
of  Conductance,  Susceptance  and  Admittance. — An  impedance 
coil  of  10  ohms  resistance  and  0.1  henry  inductance,  a  con- 
denser of  negligible  resistance  and  100  microfarads  capacitance 
and  a  non-inductive  resistance  of  20  ohms  are  connected  in  paral- 
lel across  a  200-volt,  60-cycle  circuit.  What  are  the  resultant 
current  and  power  taken  by  the  impedance  coil,  condenser  and 
resistance  in  parallel?  What  is  the  resultant  power-factor  of 
the  circuit?  What  current  and  power  does  each  branch  of  the 
circuit  take?  What  is  the  power-factor  of  each  branch? 


218  PRINCIPLES  OF  ALTERNATING  CURRENTS 

u  =  2<irf  =  377 
2!  =  10  +  J377  X  0.1 
=  10  +  J37.7  ohms. 
.      106 

22  =  0 -^377X100 

=  0  -  j'26.53  ohms. 

23  =  20  +  jO  ohms 

1  10  37.7 


2/1      21      (10)2  +  (37.7)2        (lO)2  +  (37.7)2 
=  0.006575  -  jO.02479  mho. 

1  .  26.53 

2/2  =  r  =  0  +  j 


22  '  (O)2  +  (26.53)2 

=  0  +  #.03770  mho. 
1  20  0 


23      (20)2+(0)2      J(20)2-h  (O)2 

=  0.0500  -  jO  mho. 
y0  =  SSgf  -  JSJ6 

=  0.05658  +  jO.01291  mho. 

Take  Fo,  the  voltage  drop  impressed  on  the  circuit,  as  the 
axis  of  reals.     Then 

Fo  =  200(1  +  JO) 


=  200(1  +  jO)  (0.05658  +  jO.01291) 

=  11.32  +  J2.582 
Jo  =  V(11.32)2  +  (2.5S2)2 

=  11.61  amperes. 
PO  =  FO/O  cos  00 


=  (200)  2  X  (0.05658) 
=  2263  watts. 


cos  d    = 


11  32 

jj~g?  =  0.9746  =  power-factor  for  the  circuit. 

12.95  degrees. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  219 

Since  tan  0o  =  —  is  negative  (bQ  is  negative)  ,  00  is  negative  and 
0o 

the  resultant  current  70  leads  the  voltage  V<>  impressed  on  the 
circuit  by  12.95  degrees. 

Ii  =  VQyi 

=  200(1  +  jO)  (0.006575  -  jO.02479) 

=  1.315  -  J4.958 
/i  =  \/(1.315)2  +  (4.95S)2 

=  5.13  amperes. 
Pi  =  TV0i 

=  (200)  2  X  (0.006575) 

=  263  watts. 

cos  0,  = 


1  315 
=  -^—  ^  -  =  0.2563  =  power-factor  for  branch  1. 

O.  lo 

6  1  =  75.15  degrees. 

Since  tan  0i  =  —  is  positive,  0i  is  positive  and  the  current  7i 
J7i 

lags  the  voltage  Fo  impressed  on  the  circuit  by  75.  15  degrees. 
J2  = 


=  200(1  +  jO)  (0  +  jO.03770) 
=  0  +  J7.540 


I2=  V(0)2  +  (7.540)2 
=  7.54  amperes. 

P2  =  TV<72 

=  (200) 2  X  (0) 
=  0  watts. 


cos  62  = 


=  =-^  =  0.0  =  power-factor  for  branch  2. 


=  90  degrees. 


220  PRINCIPLES  OF  ALTERNATING   CURRENTS 

Since  tan  62  =  —  is  negative,  02  is  negative  and  the  current  7  2 

02 

leads  the  voltage  Vo  impressed  on  the  circuit  by  90  degrees. 

/3  =  To£3 

=  200(1  +  jO)  (0.0500  -  JO) 
=  10.00  -  jO 

/3  =  V(io.oo)2  +  (o)2 

=  10.00  amperes. 


=  (200)  2  X  (0.0500) 
=  2000  watts 


cos  03  = 


10.00  -          ,     . 

TTTT     =  1.0  =  power  factor  for  branch  3. 


63  =  0  degrees. 

The  current  73  is  in  phase  with  the  voltage  Fo  impressed  on 
the  circuit. 

Po  =  Pi  +  P2+P3 
=  263  +  0  +  2000 
=  2263  watts. 

Parallel  Resonance.  —  A  parallel  circuit  is  said  to  be  in  reson- 
ance when  its  resultant  susceptance  is  zero. 

Io  =  V0yQ 

=  F0(<7o  -  J60)  (75) 

cos  (9o  =  —7=M=  (76) 


At  resonance  60  =  0  and 

/o  =  Fofifo  (77) 

cos  6Q  =  1  (78) 

For  fixed  resultant  conductance,  the  resultant  admittance  of  a 
parallel  circuit  is  a  minimum  at  resonance,  being  equal  to  the 
resultant  conductance.  The  resultant  current  is  also  a  minimum. 
Consequently  the  resultant  impedance  is  a  maximum  and  would 
be  infinite  if  it  were  possible  to  have  the  resultant  conductance 
zero.  For  the  resultant  conductance  to  be  zero  the  resistances  of 


IMPEDANCES  IN  SERIES  AND  PARALLEL  221 

all  parallel  branches  of  the  circuit  would  have  to  be  zero.  For  a 
series  circuit  with  fixed  resistance,  the  resultant  impedance  is  a 
minimum  at  resonance  and  is  equal  to  the  resultant  resistance. 
For  both  series  and  parallel  circuits,  the  power-factor  is  unity  at 
resonance  and  the  resultant  current  is  in  phase  with  the  voltage. 
For  a  series  circuit  in  resonance,  the  reactive  components  of  the 
voltage  drops  across  the  component  parts  of  the  circuit  add  up  to 
zero.  For  a  parallel  circuit  in  resonance,  the  reactive  components 
of  the  currents  in  the  parallel  branches  of  the  circuit  add  up 
to  zero. 

For  a  parallel  circuit  with  constant  impressed  voltage,  there 
can  be  no  rise  in  voltage  across  any  branch  of  the  circuit  at 
resonance.  Although  the  resultant  current  is  a  minimum,  the 
currents  in  the  inductive  and  capacitive  branches  may  be  large  if 
their  impedances  are  small.  Since  the  impedance  is  a  maximum 
for  parallel  resonance,  parallel  resonance  for  a  constant  current 
circuit,  i.e.,  for  a  circuit  in  which  the  current  is  maintained  con- 
stant, would  be  undesirable  in  most  cases  since  the  potential 

drop,  VQ  =  — ,  across  it  at  resonance  would  in  general  be  excessive. 

The  currents  in  the  inductive  and  capacitive  branches  probably 
would  also  be  excessive. 

In  general,  parallel  resonance  is  a  highly  desirable  condition 
for  a  constant  potential  power  circuit,  since  it  gives  a  minimum 
resultant  line  current  for  a  given  amount  of  power  transmitted. 
All  commercial  power  circuits,  except  those  for  street  lighting 
with  series  arc  or  series  incandescent  lamps,  are  operated  at 
constant  potential  and  the  loads  are  in  parallel.  A  few  series 
power  circuits,  other  than  those  for  street  lighting,  are  used 
abroad,  but  these  are  all  direct-current  circuits.  Parallel  reson- 
ance not  only  permits  the  power  to  be  transmitted  with  a  mini- 
mum line  loss,  i.e.,  with  a  minimum  72r  loss  in  the  line,  but  what 
is  more  important  it  also  permits  the  use  of  the  minimum  gen- 
erator capacity  for  a  fixed  amount  of  power  transmitted.  The 
output  of  alternating-current  generators  is  determined  by  the 
product  of  current  and  voltage,  i.e.,  by  volt-amperes,  and  not  by 
watts.  Their  output  is  therefore  a  maximum  at  unity  power- 
factor,  which  corresponds  to  resonance.  Special  devices  are 
used  in  many  cases  to  raise  the  power-factor  of  commercial 


222 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


power  circuits  by  producing  resonance  or  partial  resonance  by 
placing  capacitive  loads  in  parallel  with  the  existing  commercial 
inductive  loads.  The  saving  in  generator  and  transformer 
capacity,  line  copper  and  line  transmission  losses  will  in  many 


FIG.  64. 

cases  much  more  than  balance  the  cost  of  the  power-factor 
correcting  apparatus.  The  saving  in  space  occupied  by  the 
generating  and  distributing  apparatus  is  an  important  factor. 

Series  resonance  would  be  a  highly  desirable  condition  for  a 
constant-current  power  circuit,  since  it  would  make  the  resultant 


IMPEDANCES  IN  SERIES  AND  PARALLEL  223 

voltage  impressed  across  the  circuit  a  minimum  for  a  fixed 
amount  of  power  transmitted.  With  the  current  maintained 
constant,  there  would  be  little  danger  of  excessive  voltage  drops 
across  the  component  parts  of  the  circuit. 

Fig.  64  shows  the  effect  of  varying  the  frequency  to  produce 
parallel  resonance  in  a  circuit  consisting  of  two  branches  in 
parallel,  one  containing  an  inductance,  L  =  377  microhenrys,  in 
series  with  a  resistance,  rL  =  10  ohms,  the  other  containing  a 
capacitance,  C  =  0.00235  microfarads,  in  series  with  a  resistance, 
rc  =  10  ohms.  The  impressed  voltage  is  1  volt.  The  admit- 
tance, which  due  to  the  low  resistances  is  substantially  equal  to 
the  resultant  susceptance,  and  also  the  current  are  a  minimum  at 
resonance,  which  occurs  at  about  1.7  X  105  cycles.  The  con- 
stants of  the  circuit  for  which  Fig.  64  is  plotted  are  of  magnitudes 
such  as  might  occur  in  radio  work. 

Networks  consisting  of  series-parallel  branches  tuned  to  be 
in  series  resonance  for  one  frequency  and  in  parallel  resonance 
for  another  are  of  great  importance  in  radio  work  and  in  "carrier 
systems"  of  wire  telephony,  where  they  are  used  to  damp  out 
waves  of  one  frequency  and  to  let  through  those  of  another 
frequency.  More  will  be  said  of  this  under  "Impedances  in 
Series-parallel. " 

The  curves  in  Fig.  64  should  be  compared  with  those  for  series 
resonance  in  Fig  62,  page  207. 

Although  minimum  resultant  current  occurs  in  a  constant- 
potential  parallel  circuit  at  resonance  when  the  conductances  of 
the  parallel  branches  are  constant,  minimum  resultant  current 
does  not  occur  at  resonance  if  the  resistances  instead  of  the 
conductances  of  the  parallel  branches  are  constant.  Since 
conductance  is  a  function  of  reactance  as  well  as  of  resistance, 
when  the  resistances  of  the  parallel  branches  are  constant  the 
conductances  of  these  branches  will  not  remain  constant  when 
their  reactances  are  varied  to  produce  resonance. 

Consider  a  circuit  with  two  parallel  branches  on  which  a 
constant  voltage  at  constant  frequency  is  impressed.  Let  one 
branch  consist  of  a  constant  resistance  in  series  with  a  constant 
inductance.  Let  the  other  branch  contain  a  constant  resistance 
in  series  with  a  variable  capacitance.  The  diagram  of  connec- 
tions of  the  circuit  is  shown  in  Fig.  64a. 


223a 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Refer  to  Fig.  646.     The  current  IL  in  the  inductive  branch  is 
fixed  in  direction  and  magnitude  with  respect  to  the  voltage  V. 


V=  constant 
f=  constant 

A 


X  L-  constant 
?*/,= constant 


X  0  =  variable 
=  constant 


FIG.  64a. 


When  the  capacitance  of  the  condenser  is  infinite,  xc  is  zero  and 
the  current  Ic  in  the  condensive  branch  is  in  phase  with  the  im- 
pressed voltage  V.  Under  this  condition,  V  =  Ic\/rc2+xc2  =  Icrc. 


when  C=.A<.oc 


/  Q  when  C  = 


FIG.  646. 


If  the  capacitance  is  now  decreased,  Ic  will  decrease  and  will 
lead  the  voltage  V  which  is  still  equal  to  Ic\/rc2  +  xc2,  but  in 
this  case  xc  is  not  zero.  As  the  capacitance  is  decreased,  the  end 
of  the  vector  representing  the  Icrc  drop  will  trace  a  semicircle 
with  the  impressed  voltage  V  as  a  diameter.  This  semicircle 
is  shown  dotted  in  Fig.  646.  Since  r  is  constant,  Ic  must  vary 
directly  as  Icrc.  The  end  of  7C,  therefore,  will  also  trace  a 
semicircle.  The  diameter  of  this  semicircle  is  the  vector  7C 
for  the  condition  C  =  °c .  This  circle  is  shown  by  a  dot-and- 
dash  line.  Since  70  is  found  by  adding  to  Ic  a  vector  of  fixed 
magnitude  and  fixed  direction,  if  lc  travels  on  a  semicircle,  70 
must  also  travel  on  a  semicircle.  This  latter  semicircle  is  shown 
by  a  full  line.  Its  diameter  lies  between  the  end  of  the  vector 
70  for  the  condition  C  =  «  and  the  end  of  the  vector  IL. 

Resonance  will  occur  when  the  vector  70  is  in  phase  with  V. 
As  the  capacitance  of  the  condenser  circuit  is  decreased  from 
infinity,  resonance  will  first  occur  when  the  end  of  70  lies  at  the 
point  6.  If  the  capacitance  is  further  decreased,  70  will  decrease 


IMPEDANCES  IN  SERIES  AND  PARALLEL  2236 

and  resonance  will  again  occur  when  the  end  of  70  lies  at  the  point 
a.  Neither  of  these  resonant  conditions  corresponds  to  minimum 
IQ.  Minimum  70  will  not  occur  until  the  capacitance  is  further 
decreased  to  the  value  which  makes  the  vector  70  normal  to  the 
semicircle  c  a  6,  on  which  its  end  lies. 

When  70  has  its  minimum  value,  the  extension  of  the  vector 
which  represents  it  will  pass  through  the  center  of  the  semicircle 
cab.  A  line  drawn  between  the  end  of  the  vector  IL  and  the 
point  c  is  equal  in  magnitude  and  direction  to  the  vector  represent- 
ing the  current  taken  by  the  capacity  branch  when  the  resultant 
current  7 0  is  a  minimum .  The  conditions  which  produce  minimum 
resultant  current  can  easily  be  calculated  from  Fig.  646. 

Since  the  diameter  of  the  circle  on  which  70  swings  is  equal  to 

V  V 

—,  when  IL  sin  6  is  equal  to  s~  this  circle  will  be  tangent  to  the 

line  a6,  Fig.  646.  Under  this  condition,  there  will  be  only  one 
value  of  the  condenser  capacitance  which  will  produce  resonance. 

V 
When  LL  sin  BL  is  greater  than  — ,  resonance  cannot  be  produced 

by  any  value  of  the  condenser  capacitance. 

When  resonance  is  produced  in  power  circuits  by  use  of  an 
over-excited  synchronous  motor  which  takes  a  leading  current, 
resonance  corresponds  to  the  condition  of  minimum  resultant 
current  since  the  equivalent  conductances  of  the  loads  are  con- 
stant for  any  fixed  loads.  In  radio  circuits  resonance  produces 
practically  minimum  current  since  in  high-frequency  radio 
circuits  the  resistances  are  small  compared  with  the  reactances. 

Example  of  Parallel  Resonance. — An  impedance  of  0.1  henry 
inductance  and  10  ohms  resistance  and  a  capacitance  of  70  micro- 
farads and  negligible  resistance,  are  connected  in  parallel.  For 
what  frequency  will  the  circuit  act  like  a  non-inductive  resistance? 
What  will  be  the  value  of  this  resistance? 

XL  =  27T/L  =  6.284  X  /  X  0.1  =  0.6284  /  ohms. 
-1  -106  _  1 


For  the  circuit  to  act  like  a  non-inductive  resistance,  the  result- 
ant susceptance  of  the  circuit  must  be  zero.  For  the  susceptance 
to  be  zero,  the  circuit  must  be  in  resonance.  For  resonance 


224  PRINCIPLES  OF  ALTERNATING  CURRENTS 

26  =  0 
XL  -XC 


0.6284/ 


1430  =  100  +  0.3949/2 


1330 
X3949 

=  58.03  cycles 

XL  =  at  resonance  =  2u-  X  58.03  X  0.1 
=  36.46  ohms 


-L        rc 
10 


-I-  _   -n 

!\2       I       ff\\Z       I       xv.    2  «/ 


(10)2  +  (36.46)2       (0) 
=  0.006994  -JO 

_!_       1_ 

ZQ  ~  y0  ~  0.006994  -  jO 

=  143.0  +  JO 
r0  =  143.0  ohms. 

Impedances  in  Series-parallel. — When  a  circuit  consists  of 
a  number  of  series  elements,  some  of  which  are  made  up  of 
two  or  more  impedances  in  parallel,  the  elements  containing  the 
impedances  in  parallel  must  be  replaced  by  their  equivalent 
simple  impedances  by  means  of  equations  (70)  and  (71),  page  216. 
The  circuit  may  then  be  solved  like  a  simple  series  circuit.  An 
example  will  make  this  clear.  Two  impedances  z\  and  zz  in 
parallel  are  connected  in  series  with  a  third  impedance  z3.  How 
much  power  and  current  will  the  entire  circuit  take  when  it  is 
connected  across  200  volts?  What  will  be  the  currents  in  the 
impedances  z\  and  £2? 


IMPEDANCES  IN  SERIES  AND  PARALLEL  225 

z\  =  5  +  j2  ohms. 
22  =  6  —  j4  ohms. 

23    =    1    +  J3 

1  n  x, 


«    9       I       M    9  v*.    91       M    9 

2]          rj"  T  #l  TI     T  3Ji 

5  2 


(5)2  +  (2)2       J(V)  +  (2)2 
=  0.1723  -  jO.0690  mho. 


+  3 


(6)2  +  (4)2 
=  0.1154  +  jO.0769  mho. 

The  resultant  admittance  of  z\  and  z2  in  parallel  is 

$12  =  yi  +  £2 

=  0.2877  +  J0.0079 

=  0.2877  -  j(- 0.0079)  mho. 

The  general  form  for  admittance  is  y  =  g  —  jb.     Therefore 
612  is  negative,  as  indicated. 

The  resultant  impedance  of  the  impedances  z\  and  22  in  parallel 
is 

912  •         &i2 


12       (<7i2)2  +  (6i2)2      ''(ait)1  +  (6i2)2 

0.2877  -0.0079 

+  r. 


(0.2877) 2  +  (0.0079) 2   '   J  (0.2877) 2  +  (0.0079) 2 
3.475  -  jO.0954  ohms 


212  =  V  (3.475) 2  +  (0.0954) 2 
=  3.476  ohms. 

The  resultant  impedance  of  the  entire  circuit  is 

20    =   23    +  2i2 

=  (1  +j3)  +  (3.475  -  jO.0954) 
=  4.475  -f  J  2.905  ohms. 

15 


226  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Take  the  impressed  voltage  drop  F0  as  the  axis  of  reals.     Then 

Fo  =200+  JO 

j        Fo  =       200  +  JO 

z0      4.475  +  J2.905 

200  +  jO  4.475  -  J2.905 

4.475  +  J2.905       4.475  -  j2  905 
=  31.45  -  J20.42  amperes. 
Jo  =  \/(31.45)2  +  (20.42)2 

=  37.5  amperes. 

Po  =  total  power  =  200  X  31.45  +  0  X  (-20.42) 
=  6290  watts. 
=  JoVo 

=  (37.5)2  X  4.475 
=  6290  watts. 

The  voltage  drop  across  the  impedances  z\  and  z2  in  parallel  is 


l2    =        o 

=  (200  +  JO)  -  (31.45  -  J20.42)(l  +  J3) 
=  107.3  -  J73.93  volts. 
F12  =  V(107.3)2  +  (73.93)2 
=  130.4  volts. 

=  /ozi2  =  37.5  X  3.476  =  130.4  volts. 
130.4  130.4 


"  V(5)2  +  (2p  ~   5.39 
=  24.19  amperes. 

Fi2  =          130.4          =  130.4 

22  "  VW+~W  ~  7.21 
=  18.08  amperes. 

The  fact  that  series  resonance  gives  minimum  impedance  for 
fixed  total  resistance  and  parallel  resonance  gives  maximum 
impedance  for  fixed  total  conductance  has  already  been  men- 
tioned. This  difference  between  the  impedance  of  a  series 
circuit  and  of  a  parallel  circuit  at  resonance  is  of  great  importance 
in  radio-telegraphy  as,  by  its  use,  it  is  possible  to  so  tune  a  receiv- 
ing circuit  that  it  will  respond  to  a  definite  frequency  and  at  the 
same  time  will  suppress  an  undesirable  frequency.  This  is 


IMPEDANCES  IN  SERIES  AND  PARALLEL  227 

accomplished  by  the  use  of  what  is  known  in  wireless  work  as 
coupled  circuits.  Such  networks  consist  of  series  and  parallel 
parts  which  permit  of  independent  tuning  for  series  resonance 
for  the  desired  frequency  and  for  parallel  resonance  for  the 
frequency  to  be  suppressed.  A  simple  coupled  circuit  is  shown 
in  Fig.  65. 

L2  is  an  inductance.  Ci  and  C3  are  capacitances.  D  is  a 
detector.  The  electric  oscillations  received  by  the  antenna  are 
impressed  on  the  circuit  at  E.  With  CB  fixed,  L2  and  Ci  are 
first  adjusted  for  parallel  reson- 
ance for  the  frequency  to  be 
suppressed.  Their  resultant  par- 
allel impedance  for  this  frequency 
will,  therefore,  be  a  maximum. 
Since  the  detector  circuit  is  in 
series  with  L2  and  Ci  in  parallel,  FIQ 

little  current  of  the  frequency  to 
be  suppressed  will  flow  through  the  detector.  €3  is  now  ad- 
justed to  produce  series  resonance  in  the  series  circuit  con- 
sisting of  the  detector  and  Ca  in  series  withL2  and  C\  in  parallel. 
This  makes  the  impedance  of  the  circuit  as  a  whole  a 
minimum  for  the  frequency  to  be  detected.  By  this  double 
series-parallel  tuning,  minimum  impedance  is  produced  in  the 
detector  circuit  for  the  frequency  to  be  detected  and  maximum 
impedance  is  produced  for  the  frequency  to  be  suppressed. 

The  above  method  of  tuning  assumes  that  the  frequency  to  be 
suppressed  is  higher  than  the  one  to  be  detected.  If  the  fre- 
quency to  be  detected  were  the  higher,  C3  would  have  to  be 
replaced  by  an  inductance. 

An  Example  of  Series-parallel  Tuning  in  a  Radio  Circuit  to 
Prevent  Response  to  an  Interfering  Frequency. — The  wireless 
receiving  circuit  shown  in  Fig.  65  is  to  be  adjusted  to  respond  to 
a  frequency  of  50,000  cycles  and  to  damp  out  an  interfering 
frequency  of  100,000  cycles.  L2  is  an  impedance  of  10,000  micro- 
henrys  self-inductance  and  80  ohms  resistance.  D  is  a  ticker 
detector  having  negligible  inductance  and  15  ohms  resistance. 

The  circuit  consisting  of  L2  and  C\  in  parallel  will  first  be 
adjusted  for  parallel  resonance  for  100,000  cycles.  Then,  with- 
out changing  L2  or  Ci,  the  capacitance  Cs  will  be  adjusted  to 


228  PRINCIPLES  OF  ALTERNATING  CURRENTS 

produce  series  resonance  in  the  circuit  consisting  of  the  detector 
D  and  the  capacitance  Ca  in  series  with  the  capacitance  C\  and  the 
inductance  L2  in  parallel.  The  resistances  of  the  capacitances 
will  be  assumed  to  be  negligible. 

At  100,000  cycles 

22    =    80   +  J10-2    X    27T    X    105 

=  80  +  j6280  ohms. 
At  50,000  cycles 

z2  =  80  +  jlO-2  X  27!  X  5  X  104 
=  80  +J3140  ohms. 

For  parallel  resonance  of  L2  and  Ci  at  100,000  cycles,  the 
resultant  susceptance  of  the  circuit  consisting  of  L2  and  Ci  in 
parallel  must  be  zero.  Therefore,  since  the  resistances  of  the 
capacitances  are  assumed  to  be  negligible 


2 


9T 

co2L 

io-2 


farads 

, 

-,,  farad. 


(80) 2  +  (6280) : 
Ci  =  0.0002537  microfarad. 

At  50,000  cycles 

1  80  3140 


22       (80)2  +  (3140)2      ''(SO)2  +  (3140)2 
=  8.1  X  IO-6  -  J318  X  IO-6  mho. 

yi  =  \  =  0  +  j2ir  X  5  X  IO4  X  0.2537  X  10~9 

=  0  +J79.7  X  IO-6  mho. 

2/12  =  y\  +  #2  =  admittance  of  1  and  2  in  parallel 
=  8.1  X  IO-6  -  J238.3  X  IO-6  mho. 

212  =  _L=  142  +  J4197  ohms. 

^12 

For  series  resonance  at  50,000  cycles  for  the  entire  circuit,  the 
resultant  reactance  of  the  entire  circuit  must  be  zero.     Therefore 

TT-  =  4197  ohms 


4197  X  27T  X  50,000 
=  0.0007585  microfarad. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  229 

Total  impedance   of   the   entire   circuit   at   50,000   cycles   is 
20  =  (15  +  JO)  +  (142  +  #197)  +  (0  -  #197) 

=  157  +  jO  ohms. 
z0  =  157  ohms. 

At  100,000  cycles. 

J.  _  80  6280 

:    2z  ~  (80)  2>  (6280)  2       J  (80)  2  +  (6280)  2 
=  0.00000203  -  jO.0001594  mho. 
yj  =  J27T  X  105  X  0.0002537  X  10~6 
=  jO.0001594  mho. 

2/12  =  2/i  +  fa 

=  2.03  X  10-6  -  jO  mho. 


_  . 

yl2       2.03  X  10-6  -  JO 
=  492,000  +  jO  ohms. 


3  =      ~~    2*  X  105  X  0.7585  X  10~9 

=  0  -  J2100  ohms. 
zo  =  (15  +  JO)  +  (492,000  +  JO)  +  (0  -  J2100) 

=  492,000  -  J2100 
z0  =  492,000  ohms. 

The  ratio  of  the  impedances  of  the  entire  circuit  at  the  two 
frequencies  is 

go  (at  100,000  cycles)  =  492,000 
z0(at    50,000  cycles)  "       157 
=  3130 

If  the  interfering  waves  and  the  waves  for  which  response 
is  desired  were  of  the  same  intensity,  the  current  in  the  receiving 

circuit  due  to  the  interfering  frequency  would  be  only  ^-—  of  that 


caused  by  the  waves  of  the  frequency  to  be  detected. 

Filter  Circuits.  —  The  principle  of  the  so-called  filter  circuits, 
which  are  used  in  wireless  telephony  and  also  in  the  carrier-wave 
system  of  wire  telephony  to  filter  out  waves  of  certain  frequency, 
are  nothing  more  than  a  combination  of  series  and  parallel 
circuits  that  can  be  tuned  for  both  series  and  parallel  resonance 
to  suppress  waves  of  a  certain  frequency  and  to  allow  waves  of 
another  frequency  to  pass. 


230  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Jn  the  carrier  system  of  wire  telephony,  speech  is  transmitted 
by  very  high  frequency  currents,  which  are  above  audible 
frequency.  The  intensity  of  these  high  frequency  waves  is  modu- 
lated, i.e.,  varied  to  correspond  to  the  variation  in  intensity 
produced  in  the  air  waves  by  the  voice.  Several  of  these  high 
frequency  carrier  waves,  each  of  a  different  frequency,  are 
superposed  on  a  single  circuit.  At  the  receiving  end  of  the  line 
each  wave  is  filtered  out  and  delivered  to  the  proper  receiving 
circuit.  At  the  present  time  (1921)  four  two-way  carrier- wave 
conversations,  i.e.,  four  each  way,  may  be  carried  on  over  a 
single  circuit,  in  addition  to  the  ordinary  telephonic  and  tele- 
graphic messages. 

Effect  of  Change  of  Reactance  and  Resistance  with  Current. 
Reactance  and  Resistance  Functions  of  Current. — In  all  the 
equations  involving  resistance  or  reactance  or  both,  which  have 
been  considered  thus  far,  both  resistance  and  reactance  have  been 
assumed  constant.  The  equations  are  true  only  under  these 
conditions. 

When  a  sinusoidal  voltage  is  impressed  on  a  circuit  of  constant 
impedance,  the  resultant  current  under  steady  conditions  is  also 
sinusoidal.  When,  however,  either  the  resistance  or  the  reac- 
tance of  the  circuit  is  not  constant  but  varies  with  the  current, 
i.e.,  is  a  function  of  the  current,  a  sinusoidal  voltage  impressed  on 
the  circuit  will  give  rise  to  a  non-sinusoidal  current. 

The  inductance  of  a  circuit  containing  iron  cannot  be  constant, 
since  the  flux  is  not  proportional  to  the  current.  The  flux  per 
ampere  is  a  variable.  The  reactance  of  such  a  circuit  at  any  fixed 
frequency  is  a  function  of  the  current,  except  at  very  low  flux 
densities  when  it  may  be  approximately  constant. 

The  resistance  of  a  circuit,  which  has  so  small  a  heat  capacity 
that  the  variation  in  the  72r  loss  during  a  cycle  produces  an 
appreciable  change  in  temperature,  must  vary  with  the  current 
during  each  cycle,  unless  the  coefficient  of  resistance  variation 
with  temperature  is  zero.  If  the  temperature  coefficient  is  posi- 
tive like  that  of  a  metal,  a  sinusoidal  voltage  will  produce  a 
current  wave  which  is  flatter  than  a  sine  wave.  The  current  will 
contain  harmonics.  In  such  a  case  the  power-factor  of  the  circuit 
cannot  be  unity,  even  though  the  circuit  contain  no  reactance, 


IMPEDANCES  IN  SERIES  AND  PARALLEL  231 

since  the  product  of  current  and  voltage  cannot  be  equal  to 
the  true  power  absorbed. 

The  average  power  absorbed  in  heating  by  a  circuit  containing 
a  resistance  which  is  a  function  of  the  current  is 

•T 
i*rdt 


.f 

Tjo 

-If 

Tjo 


(79) 


This  integral  can  be  equal  to  (7r.m.«.)2r  only  when  r  is  inde- 
pendendent  of  the  current  strength.  The  apparent  resistance  of 
a  circuit  is  always 

r  =  p  (80) 

where  P  is  the  average  power  absorbed  due  to  the  resistance  and 
I  is  the  root-mean-square  value  of  the  current.     (See  Effective 
Resistance,  page  234.) 
The  expressions 

p  v 

r  =  TO  and  r  =  - 
i2  i 

where  r,  p  and  i  are  instantaneous  values  of  resistance,  power  and 
current  and  v  is  the  instantaneous  voltage  drop  due  to  resistance, 
are  always  equal. 

In  practice  the  change  in  temperature  of  a  conductor  with 
current  during  a  cycle  is  usually  too  small  to  produce  appreciable 
effect.  Low  candle-power,  high-voltage,  incandescent  lamps, 
however,  especially  those  like  the  tungsten  lamp  with  extremely 
fine  filaments,  do  show  an  appreciable  variation  in  resistance 
with  current  during  a  cycle  even  at  frequencies  as  high  as  sixty 
cycles. 

If  a  circuit  contains  iron,  there  may  be  a  very  appreciable 
change  in  its  inductance  with  current  during  each  cycle,  especially 
when  the  iron  is  worked  at  high  saturation.  Due  to  this  varia- 
tion, marked  harmonics  may  be  present  in  the  current  which  are 
not  present  in  the  impressed  voltage.  With  a  sinusoidal  voltage 
impressed  on  the  circuit,  a  third  harmonic  with  a  maximum 
value  as  great  as  25  to  40  per  cent,  of  the  maximum  value  of  the 
fundamental  would  not  be  excessive.  A  third  harmonic  of  this 
magnitude  often  occurs  in  the  no-load  current  of  a  transformer 
with  sinusoidal  impressed  voltage, 


232  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  voltage  drop  across  a  circuit  containing  constant  resis- 
tance and  constant  inductance  is  given  by 

e  =  ri  +  LAi  (81) 

If  L  =  N-J-.    decreases  with  increase  of  current,  as  it  must 

when  iron  is  present,  then  for  a  given  impressed  voltage,  -r  must 

di 

increase  more  rapidly  than  when  L  is  constant  in  order  that  L-r 

plus  ri  shall  be  equal  to  the  impressed  voltage  at  each  instant. 
A  sinusoidal  voltage  impressed  on  a  circuit  containing  iron, 
therefore,  will  give  rise  to  a  current  which  will  be  more  peaked 
than  a  sinusoid. 

Non-sinusoidal  Voltage  Impressed  on  a  Circuit  Containing 
Constant  Impedances  in  Series  and  in  Parallel. — When  a  non- 
sinusoidal  voltage  is  impressed  on  a  circuit  consisting  of  constant 
resistances,  constant  inductances  and  constant  capacitances  in 
series  or  in  parallel,  the  fundamental  and  each  harmonic  must  be 
considered  separately.  They  cannot  be  combined  until  the  final 
solution  is  reached  when  the  root-mean-square  value  of  the 
current  and  voltage  drops  may  be  found  in  the  usual  manner  by 
taking  the  square  root  of  the  sum  of  the  squares  of  the  compon- 
ents caused  by  the  fundamental  and  each  harmonic  acting 
separately.  It  must  be  remembered  that  reactance,  susceptance 
and  conductance  are  functions  of  frequency.  For  constant  resis- 
tance, inductance  and  capacitance,  inductive  reactance  is  propor- 
tional to  frequency  and  capacitive  reactance  is  inversely  pro- 
portional to  frequency.  Conductance  decreases  with  increase  in 
frequency  for  a  circuit  containing  inductance  and  resistance  only 
and  increases  with  increase  in  frequency  for  a  circuit  containing 
capacitance  and  resistance  only.  Conductance  must  be  found 
for  each  frequency  from  the  resistance,  inductance  and  capaci- 
tance. Susceptance  may  either  increase  or  decrease  with  fre- 
quency, depending  on  the  relative  magnitudes  of  resistance  and 
reactance  and  the  way  the  reactance  changes  with  frequency. 
Susceptance,  like  conductance,  must  be  found  for  each  frequency 
from  the  resistance,  inductance  and  capacitance. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  233 

Example  of  a  Simple  Series  Circuit  on  which  a  Non-sinusoidal 
Voltage  is  Impressed. — A  voltage  whose  equation  is 

e0  =  300  sin  314*  +  75  sin  942* 

is  impressed  on  a  circuit  consisting  of  a  condenser  of  negligible 
resistance  and  25  microfarads  capacitance  in  series  with  an 
impedance  of  40  ohms  resistance  and  0.1  henry  inductance. 
What  is  the  equation  of  the  current  and  what  is  the  root-mean- 
square  value? 

The  impressed  voltage  contains  a  fundamental  of 


and  a  harmonic  of 


/  =  -^—  =  50  cycles 

ZTT 

942 
/  =  -—  =  150  cycles. 


It  therefore  contains  a  fundamental  and  a  third  harmonic. 
For  the  fundamental 

Eml 


300 


=  _  300  __  =    300 
"  \/(40)2  +  (31.4  -  127.4)2  ~  104.0 
=  2.884  amperes. 

31.4  -  127.4       -96.0 
-  -     --    -  =  --  =  -2.40 


0i  =  -67.38  degrees. 
For  the  third  harmonic 


75 


75 =    75 

V(40)2  +  (94.2  -^4&5p       65.4 
1.146  amperes. 


234  PRINCIPLES  OF  ALTERNATING  CURRENTS 

KI  7 
tan  B3  =  j^-  =  1.293 

63  =  52.3  degrees. 

to  =  2.88  sin  (314*  +  67?4)  +  1.15  sin  (942*  -  52?3) 

j^  s   =     /^88)~ 


2 
=  2.19  amperes. 

Effective  Resistance. — When  a  direct  current  flows  in  a 
circuit,  there  is  a  loss  in  power  due  to  the  resistance  of  the  circuit. 
This  loss  is  the  so-called  copper  loss  and  is  equal  to  the  current 
squared  multiplied  by  the  ohmic  resistance  of  the  circuit.  The 
resistance  is  equal  to  the  copper  loss  divided  by  the  current 
squared.  Similarly,  when  an  equal  alternating  current  flows  in 
the  same  circuit  there  is  also  a  loss  in  power,  but,  in  general,  the 
loss  is  greater  with  alternating  current  than  with  direct  current. 
On  account  of  this  increase  in  loss,  the  apparent  resistance  of  the 
circuit  is  greater  with  alternating  current  than  with  direct 
current,  and  under  certain  conditions  it  may  be  many  times 
greater.  The  apparent  resistance  of  a  circuit  with  alternating 
current  is  always  equal  to  the  loss  in  power  caused  by  the  current 
divided  by  the  current  squared.  The  resistance  of  a  circuit 
with  direct  current  is  not  only  equal  to  the  loss  caused  by  the 
current  divided  by  the  current  squared,  but  it  is  also  equal  to  the 
drop  in  potential  across  the  circuit  caused  by  the  current  divided 
by  the  current.  The  apparent  resistance  of  a  circuit  with  alter- 
nating current  does  not  equal  the  drop  in  potential  across  it 
divided  by  the  current,  except  in  very  special  cases. 

The  increase  in  loss  with  alternating  current  is  due  to  two 
causes:  first,  to  certain  local  losses  produced  by  the  varying 
magnetic  field  in  the  surrounding  material  and  in  the  conductor 
itself,  and  second,  to  the  non-uniform  current  density  over  the 
cross  section  of  the  conductor.  With  direct  current,  the  current 
density  is  uniform  over  the  cross  section  of  ordinary  conductors, 
i.e.,  conductors  of  a  single  material  of  uniform  specific  resistance. 
The  local  losses  produced  by  an  alternating  current  are:  eddy- 
current  and  hysteresis  losses  in  adjacent  magnetic  material, 
eddy-current  losses  in  adjacent  conducting  material,  and  eddy- 
current  loss  and  also  hysteresis  loss,  if  the  conductor  is  magnetic, 
in  the  conductor  itself. 


IMPEDANCES  IN  SERIES  AND  PARALLEL  235 

When  an  alternating  current  flows  in  a  circuit,  the  power 
absorbed  is  equal  to  the  current  multiplied  by  the  active  compo- 
nent of  the  voltage  drop  produced  by  the  current.  Any  increase 
in  the  active  component  of  the  voltage  drop,  due  to  local  losses 
produced  by  the  current,  is  equivalent  to  an  increase  in  the 
apparent  resistance  of  the  circuit.  The  active  component  of  the 
voltage  drop  is  increased  by  all  local  eddy-current  and  hysteresis 
losses  caused  by  the  current  and  also  by  any  distortion  of  the 
current  distribution  over  the  cross  section  of  the  conductor.  The 
distortion  of  the  current  distribution  has  much  the  same  effect, 
so  far  as  the  resistance  of  the  conductor  is  concerned,  as  decreas- 
ing the  cross  section  of  the  conductor.  It  thus  increases  its 
apparent  resistance.  The  apparent  resistance  of  a  circuit  to 
an  alternating  current  is  known  as  its  effective  resistance.  Effec- 
tive resistance  is  not  a  constant.  It  varies  with  frequency,  and 
also  with  current  strength  if  the  circuit  is  adjacent  to  magnetic 
material.  In  general,  effective  resistance  increases  rapidly  with 
increase  of  frequency.  In  all  work  dealing  with  alternating 
currents,  effective  resistance  must  be  used  in 
finding  the  I2r  loss  and  the  potential  drop  in 
the  circuit  due  to  resistance. 

The  non-uniform  distribution  of  the  current 
over  the  cross-section  of  a  conductor  is  due 
to  the  difference  in  the  reactances  of  ele- 
ments of  the  conductor  taken  parallel  to 

rIG.    DO. 

its    axis.     Let  Fig.   66   represent   the    cross- 
section  of  a  conductor.     Consider  any  circular  element  in  the 
conductor  of  radius  x  and  width  dx. 

The  flux  lines  caused  by  a  current  in  a  straight  cylindrical 
conductor,  which  lies  in  a  medium  of  uniform  permeability,  are 
concentric  circles  with  their  centers  on  the  axis  of  the  conductor. 
These  flux  lines  not  only  surround  the  conductor  but  exist  in 
the  conductor  itself.  The  field  intensity  at  a  point  outside  any 
cylindrical  element,  such  as  the  one  shown  in  Fig.  66,  due  to  the 

27 

current  in  the  element  is  equal  to  -^  /*,  where  /  is  the  current  in 

the  element,  R  is  the  perpendicular  distance  between  the 
point  and  the  axis  of  the  conductor  and  /*  is  the  permeability  of 
the  medium  at  the  point  considered.  The  field  intensity  inside 


236  PRINCIPLES  OF  ALTERNATING  CURRENTS 

the  element  due  to  the  current  it  carries  is  zero.  Since  the  only 
flux  which  can  link  any  element,  such  as  the  one  of  radius  x 
shown  in  the  figure,  is  that  which  lies  without  it,  it  is  evident  that 
the  flux  linking  an  element  must  increase  as  its  radius  decreases. 
All  the  flux  produced  by  the  entire  conductor  links  the  element 
at  its  center.  This  includes  the  flux  within  as  well  as  without  the 
conductor.  The  only  flux  that  can  link  an  element  at  the  surface 
of  the  conductor  is  the  flux  which  lies  without  the  conductor. 

The  self -inductance  per  unit  axial  length  of  any  element  is 
equal  to  the  flux  linking  it  per  unit  length  per  unit  current.  This 
is  equal  to  all  that  flux  outside  the  element  which  is  contained 
between  two  planes  drawn  perpendicular  to  the  axi's  of  the  con- 
ductor at  unit  distance  apart.  It  is  obvious,  therefore,  that  the 
self-inductance  of  the  elements  will  increase  with  decrease  in 
their  radii,  will  be  least  for  the  element  at  the  surface  of  the  con- 
ductor and  greatest  for  the  element  at  the  center.  The  reac- 
tance will  therefore  be  least  for  the  element  at  the  surface  and 
will  be  greatest  for  the  element  at  the  center.  Since  reactance 
(27T/L)  is  proportional  to  frequency,  the  difference  between  the 
reactance  of  an  element  at  the  center  and  at  the  surface  of  any 
conductor  of  fixed  radius  will  increase  with  frequency.  It  will 
also  increase  with  the  radius  of  the  conductor  and  its  permea- 
bility. It  will  be  much  greater  for  iron  conductors  than  for 
those  of  copper. 

The  magnitude  of  the  current  in  any  element  of  a  conductor 
will  be  inversely  as  the  impedance  of  the  element.  The  current 
will  therefore  be  greatest  in  elements  at  the  surface  and  least  in 
elements  at  the  center.  At  very  high  frequencies  very  little 
current  will  flow  through  the  central  portion  of  a  large  conductor. 
Nearly  all  of  the  current  will  be  carried  by  the  portion  near  its 
surface.  This  crowding  of  an  alternating  current  towards  the 
surface  of  a  conductor  is  known  as  skin  effect.  It  is  small  for 
conductors  of  small  cross  section  carrying  low  frequency  currents, 
but  for  large  conductors  or  high  frequencies  it  may  become  very 
great.  At  very  high  frequencies,  such  as  are  used  for  radio- 
communication,  the  resistance  of  a  cylindrical  tube  may  be  very 
nearly  as  low  as  the  resistance  of  a  solid  conductor  of  equal  radius 
and  the  same  material. 

The  skin  effect  can  be  calculated  for  non-magnetic  conductors 


IMPEDANCES  IN  SERIES  AND  PARALLEL 


237 


of  circular  section  which  are  not  adjacent  to  other  conductors. 
For  conductors  of  non-circular  section  it  must  be  determined 
experimentally.  In  certain  simple  cases,  when  conductors  are 
embedded  in  iron,  as  are  the  armature  conductors  of  an 
alternator,  it  is  possible  to  calculate  the  skin  effect  with  a  fair 
degree  of  approximation. 

The  skin  effect  at  60-cycles  for  straight,  stranded,  copper  wires 
of  circular  cross  section  is  shown  in  Fig.  67. 


150 

1-10 

\m 

100 

/ 

Skin  Effect 
at 
60  Cycles 
for 
Stranded  Copper  Cable 

/ 

/ 

'          / 

/ 

/ 

/ 

/ 

"0       500,000  1,000,000  1,500.0002,000,0002,500,0003,000,000 
Circular  Mils 

FIG.  67. 

In  general,  the  current  density  in  a  conductor  is  least  at  the 
points  where  the  flux  linkages  per  ampere  due  to  the  current  in 
the  conductor  are  greatest. 

The  apparent  or  effective  resistance  -of  a  circuit  not  containing 
a  source  of  electromotive  force  may  be  found  by  measuring  the 
power  absorbed  when  a  known  current  of  definite  frequency  is 
passed  through  it.  If  P  is  the  power  absorbed  due  to  the  current 
7  at  a  frequency  /,  the  effective  resistance  of  the  circuit  at  the 
frequency  /  is 

r,  =  p  (82) 

Effective  Reactance. — The  effective  reactance  of  a  circuit  to 
alternating  current,  especially  if  there  be  iron  present,  is  always 
less  than  the  true  reactance.  The  true  reactance  is  equal  to 
2wfL  where  L  is  the  self-inductance  of  the  circuit  and  /  is  the 
frequency.  The  self -inductance,  L,  is  equal  to  the  flux  linkages 


238  PRINCIPLES  OF  ALTERNATING  CURRENTS 

of  the  circuit  per  unit  current  producing  the  flux.  The  entire 
current  in  a  circuit  is  not  effective  in  producing  flux  except  when 
no  eddy-current  or  hysteresis  losses  are  caused  by  it. 

The  power  absorbed  in  any  circuit,  exclusive  of  that  due 
to  copper  loss,  is  always  equal  to  El  cos  0,  where  E  and  I  are 
the  back  electromotive  force  and  the  current  respectively.  The 
angle  0f  is  the  phase  angle  between  E  and  7.  If  there  is  no 
power  absorbed  by  the  circuit  other  than  the  true  ohmic  cop- 
per loss,  the  angle  0f  between  E  and  I  must  be  90  degrees  and 
E  and  I  will  be  in  quadrature.  Under  this  condition,  El  cos0f 
=  0  and  VI  cos  0f  =  Pr0?, .  is  the  true  copper  loss,  where  V  is 
the  voltage  drop  impressed  across  the  entire  circuit  and  r0h.  is 
the  ohmic  resistance  of  the  circuit. 

If  there  be  magnetic  or  conducting  material  near  the  circuit,  in 
which  hysteresis  or  eddy-current  losses  occur,  the  current  must 
have  a  component  in  phase  opposition  to  the  voltage  rise 
induced  by  the  flux,  to  supply  these  losses.  This  component 
does  not  contribute  to  the  flux.  Its  effect,  so  far  as  flux  is  con- 
cerned, is  canceled  by  the  reaction  due  to  the  eddy-current  and 
hysteresis  losses.  The  only  component  of  the  current  which  is 
effective  in  producing  flux  is  that  in  phase  with  the  flux  and, 
therefore,  in  quadrature  with  the  voltage  induced  by  the  flux. 
This  component  is  equal  to  the  current  that  would  be  required  to 
produce  the  flux  were  there  no  eddy-current  or  hysteresis  losses. 
It  is  called  the  magnetizing  component  of  the  current  or  simply 
the  magnetizing  current. 

Consider  a  coil  of  wire  wound  on  an  iron  core.  If  an  alternat- 
ing current  is  passed  through  the  coil,  a  flux  will  be  set  up  which 
will  link  the  turns  of  the  coil.  The  current,  as  has  been  stated, 
will  not  be  in  phase  with  this  flux  if  there  are  eddy-current  or 
hysteresis  losses.  Let  L  be  the  flux  linkages  with  the  circuit  per 
unit  current  producing  the  flux,  i.e.,  per  unit  of  the  component  of 
current  in  phase  with  the  flux.  Call  this  component  1^.  Then 
the  induced  voltage  drop  through  the  coil,  caused  by  its  true 
self-inductance  L,  is  2-jrfLI^. 

Let  p,  Fig.  68,  be  the  flux  produced  by  the  current  7  and 
let  E  =  27r/L7^  be  the  voltage  drop  caused  by  this  flux,  where  7^ 
is  the  component  of  the  current  7  in  phase  with  the  flux  <p.  The 


IMPEDANCES  IN  SERIES  AND  PARALLEL  239 

flux  and  voltage  drop  must  be  in  quadrature,  with  the  voltage 
drop  leading  the  flux.     This  phase  relation  follows  from 

T  di 

Gdrop   —  Li^r- 

If  the  current  is  a  sine  function  of  time,  the  voltage  drop  edrop, 

which  is  proportional  to  the  derivative  of  the  current  with  respect 

to  time,  must  be  a  cosine  function  of 

time,  and  since  the  cosine  of  an  angle 

is  equal  to  the  sine  of  the  angle  plus 

ninety   degrees,    the  drop  must  lead 

the  current  by  ninety  degrees. 

The  drop  due  to  ohmic  resistance 
is  in  phase  with  the  current.  Lay 
this  off  as  shown.  V  is  the  total 
voltage  drop  across  the  circuit  and  is  FIQ  6g 

equal  to  the  true  self-induced  voltage 

drop,  2T/L/^,  plus  the  ohmic  resistance  drop,  Jr.     The  effective 
resistance  is 

P  _  VI  cos  6  _  ac  on  figure 

Te~P~   ~P~          ~T~ 

The  true  ohmic  resistance  drop  is  oh.  The  distance  ac  is 
the  active  component  of  the  voltage  drop  with  respect  to  the  cur- 
rent. This  is  the  apparent  resistance  drop.  Similarly,  ob  = 
2-jrfLIr  is  the  true  reactance  drop.  The  distance  oc  is  the  reactive 
component  of  the  voltage  drop  with  respect  to  the  current. 
It  is  in  quadrature  with  the  total  current  7  and  is  the  apparent  or 
effective  reactance  drop.  The  apparent  or  effective  reactance 
therefore  is  equal  to  the  reactive  component  of  the  voltage  drop 
across  the  circuit  divided  by  the  total  current.  The  true 
reactance  is  equal  to  the  vector  difference  of  the  voltage  drop 
across  the  circuit  and  the  true  ohmic  resistance  divided  by  the 
magnetizing  component  of  the  current,  i.e.,  by  the  component  of 
the  current  in  quadrature  with  the  vector  difference  of  the  voltage 
drop  across  the  circuit  and  the  true  ohmic  resistance  drop. 

m  0  f,       ob  on  figure 

True  reactance  x  =  2irfL  =  - 

*9 

TT, «.    x-  oc  on  figure 

Effective  reactance  xf  —  -     — v 

Effective  impedance  ze  —   ,  =  \/r<?  +  xe2 


240  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  effective  reactance  is  always  less  than  the  true  reactance 
and  the  effective  resistance  is  always  greater  than  the  true  re- 
sistance. The  differences  are  greatest  when  the  local  losses 
caused  by  the  current  are  greatest.  For  circuits  in  which  those 
losses  are  small,  the  effective  reactance  may  be  very  nearly  equal 
to  the  true  reactance  and  the  effective  resistance  may  be  very 
nearly  equal  to  the  true  resistance. 

In  general,  if  a  circuit  containing  resistance  and  reactance 
takes  P  watts  and  /  amperes  at  V  volts 

p 

re  (effective)  =    ^ohms,  (83) 


y 

ze  (effective)  =  y  ohms,  (84) 

xe  (effective)  =  \/ze*  —  re2  ohms,  (85) 

p 
ge  effective)  =  ™j  mhos,  (86) 

ye  (effective)  =  -^  mhos,  (87) 

be  (effective)  =  \/ye2  —  ge2  mhos.  (88) 

Equivalent  Resistance,  Reactance  and  Impedance  of  a  Circuit 
and  also  Equivalent  Conductance,  Susceptance  and  Admittance 
of  a  Circuit.  —  It  is  frequently  convenient,  when  dealing  with 
certain  problems,  to  replace  an  actual  load,  which  may  consist  of  a 
loaded  motor  or  any  other  load,  by  an  impedance  or  admittance 
which  will  take  the  power  at  the  same  current  and  power-factor 
as  the  actual  load.  So  far  as  the  conditions  existing  in  the 
circuit  are  concerned,  the  impedance  or  admittance  will  exactly 
replace  the  actual  load.  Since  the  current,  power  and  power- 
factor  of  commerical  loads,  such  as  motor  loads,  are  continually 
changing,  the  apparent  impedance  or  admittance  of  a  load 
cannot,  in  general,  be  constant. 

The  equivalent  constants  of  any  load  may  be  found  from  equa- 
tions (83)  to  (88)  inclusive  by  substituting  for  P,  I  and  V  the 
power,  current  and  voltage  of  the  load. 

Example  of  the  Use  of  the  Equivalent  Constants  of  a  Load.  — 
An  induction  motor  and  a  synchronous  motor,  each  rated  at 


IMPEDANCES  IN  SERIES  AND  PARALLEL  241 

2300  volts,  are  to  be  operated  in  parallel  at  the  end  of  a  trans- 
mission line  having  a  resistance  rL  =  1.0  ohm  and  an  inductive 
reactance  XL  =  1.1  ohms.  The  average  power  and  current 
taken  by  each  motor  when  operated  at  rated  voltage  with  the 
particular  load  it  has  to  carry  are: 


Induction  motor 


Synchronous  motor 


i  =  2300  volts. 
i  =    111.5  amperes. 
Pt  =    200  kilowatts 

V8  =  2300  volts. 
Is  =     100.0  amperes. 
Ps  =     150  kilowatts. 


What  must  be  the  voltage  at  the  power  station  end  of  the  line 
in  order  to  maintain  2300  volts  at  the  motors  when  they  are 
operating  under  average  load  conditions?  What  is  the  resultant 
power-factor  of  the  load  measured  at  the  motors?  What  is  the 
resultant  power-factor  measured  at  the  station  end  of  the  line. 
An  induction  motor  always  takes  a  lagging  current.  The  ex- 
citation of  the  synchronous  motor  is  adjusted  so  that  it  takes 
a  leading  current. 

The  equivalent  constants  of  the  motor  loads  are: 
Induction  motor. 

*  -  ^  -  °-0485  mh°- 

200  X  1000 


=  V(0.0485)2  -  (0.0378)  2  =  0.0304   mho. 
Synchronous  motor. 

100.0 
V*  =  =  0.0435  mho. 


gs  =     723Q0)2        =  0.02835  mho. 
b,  =  vV  -  gS 

=  V  (0.0435) 2  -  (0.02835) 2  =  -0.03295  mho. 

;,6 


242  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  equivalent  susceptance  ba  of  the  synchronous  motor  is 
negative  since  the  synchronous  motor  takes  a  leading  current. 

go  =  Qi  +  gs  =  0.0378  +  0.02835 

=  0.06615  mho. 
60  =  6t  +  bs  =  0.0304  -  0.03295 

=  -0.00255  mho. 
o  0.06615  0.06615 


r°  ~ 


+  &o2       (0.06615) 2  +  (0.00255) 2       0.004385 

=  15.08  ohms. 
60  -0.00255  -0.00255 


go2  H-  602       (0.06615)2  +  (0.00255)2  ~     0.004385 

=  -0.582  ohm. 

2/o  =  V (0.06615) 2  f-  (0.00255) 2 
=  0.06625  mho. 

The  resultant  power-factor  of  the  load,  measured  at  the  motors 
is 


p.f.  (at  motors)  -       =  =0.998 


The  current  taken  by  the  two  motors  in  parallel  is 

/o  =  7o!/o 

=  2300  X  0.06625 
=•  152.4  amperes. 

This  current  must  lead  the  voltage  F0  impressed  on  the  motors 
in  parallel  since  tan  00  =  —  =  -  ^-7^^^-  is  negative. 

QQ  U.UbolO 

The  voltage  VQ   at  the  station  which  will  give  2300  volts  at 
the  motors  is 


+  rja  +  (x0  +  XL)* 
=  152.4V(15.08  +  I)2  +  (-0.568  +  l.l)2 
=  152.4  V(16.08)2  +  (0.532)  2 
=  152.4  X  16.09 
=  2452  volts. 


IMPEDANCES  IN  SERIES  AND  PARALLEL 
The  power-factor  at  the  station  is 
p.  f.  (at  station)  = 


243 


+  rL 


V  (T*O  +  rL)2  +  (XQ  -f  xL)2 
=  16.08 
=  16.09 
=  0.999 

The  current  lags  the  voltage  at  the  station  since  tan  Bt 


-f-  XL  . 


is  positive. 


General  Summary  of  the  Conditions  in  Series  and  Parallel 

Circuits.— 

Parallel 

I2  +  ?!  + etc- 

#2  =  #3  =  etc. 


Series 
7o  =  7i  =  7s  =  7s_  =  etc. 

Fo  =  #1  +  #2  +  #3  +  etc. 
r0  =  n  +  r2  +  r3  +  etc. 
TO  =  zi  +  xj,  +  x3  +  etc. 

20    =   f  0   +  J 


Fo  =Ei 

go  =  g\  +  02  +  03  +  etc. 

b0  =  6,  +  62  + 
yo  =  go  —  jbQ 


2's  cannot  be  added  directly. 


+  etc. 
y's  cannot  be  added  directly. 


Power  factor  =  cos00  = 


V  +  x02 

Fo  =  7oZo  =  7o(r0  +  jxo)     is     an 
equation  of  voltage. 

7ofo  =  active  component  of  volt- 
age drop  with  respect  to 
current. 


Power  factor  =  cos  0o   = 


7o  =  Fo#o  =  Eo(go  -  jbo)   is  an 

equation  of  current. 
Voffo  —  active  component  of  cur- 
rent with  respect  to  volt- 
age drop. 

reactive  component  of  volt-  — /Fo&o  =  reactive  component  of  cur- 
age  drop  with  respect  to  rent  with  respect  to  volt- 
age drop. 

Power  =  Po  =  F020o 


current. 
Power  =  Po  =  /o 


60 


7o 

YQ 

/o 
Fo 


X0 

ro2  + 


244 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Series  Resonance 
For  series  resonance 

2x  =  X0  =  0 

The  resultant  impedance  for  fixed 
resistance  is  a  minimum  and  for 
fixed  impressed  voltage  the  current 
is  a  maximum. 

With  fixed  impressed  voltage  the 
voltage  drops  across  the  inductance 
and  capacitance  may  be  excessive. 
If  the  resistance  is  low  compared 
with  the  inductive  and  capacitive 
reactances,  the  voltage  drops  across 
the  inductive  and  capacitive  reac- 
tances may  be  dangerously  large. 

For  fixed  current  and  fixed  re- 
sultant resistance  and  therefore  for 
fixed. power  (P0  =  /o2r0)  the  result- 
ant voltage  drop  across  the  entire 
circuit  is  a  minimum.  This  is  a 
very  desirable  condition  for  a  con- 
stant-current power  circuit. 


Parallel  Resonance 
For  parallel  resonance 
26  =  60  =  0 

The  resultant  admittance  for  fixed 
conductance  is  a  minimum  (the  re- 
sultant impedance  is  a  maximum)  and 
for  fixed  impressed  voltage  the  cur- 
rent is  a  minimum. 

With  constant  resultant  current  the 
currents  in  the  inductive  and  capaci- 
tive branches  may  be  excessive. 
There  can  be  no  dangerously  high 
voltages  in  any  part  of  the  circuit.  If 
the  resistances  of  parallel  inductive 
and  capacitive  branches  are  very 
small,  the  currents  in  the  branches 
may  be  very  large. 

For  fixed  voltage  and  fixed  result- 
ant conductance  and  therefore  for 
fixed  power  (P0  =  V0*go)  the  resultant 
current  in  the  circuit  is  a  minimum. 
This  is  a  very  desirable  condition  for 
a  constant-potential  power  circuit. 


CHAPTER  VIII 

POLYPHASE  CURRENTS 

Generation  of  Polyphase  Currents. — A  polyphase  alternator 
differs  from  a  single-phase  alternator  only  in  the  number  of  its 
armature  windings.  A  single-phase  alternator  has  a  single 
armature  winding.  A  polyphase  alternator  has  as  many  inde- 
pendent armature  windings  as  there  are  phases.  These  windings 
are  displaced  from  one  another  by  equal  angles,  the  angles  being 
determined  by  the  number  of  phases.  If  a  two-pole  alternator 
had  two  independent  armature  windings  displaced  ninety  degrees 
from  each  other,  the  coil  sides  of  one  winding  would  lie  under  the 
centers  of  the  poles  when  the  coil  sides  of  the  other  winding  were 
midway  between  the  poles.  The  voltages  induced  in  the  two 
windings  therefore  would  be  in  time  quadrature  or  90  degrees 
apart  in  time-phase.  If  the  terminals  of  the  windings  were 
brought  out  to  insulated  slip-rings  mounted  on  the  armature 
shaft,  two-phase  currents  could  be  taken  from  the  alternator. 
If  the  alternator  had  three  independent  armature  windings, 
displaced  120  degrees  from  one  another,  the  voltages  induced  in 
them  would  be  120  degrees  apart  in  time-phase.  Three-phase 
voltages  would  be  generated  and  the  alternator  would  be  a  three- 
phase  alternator.  In  general,  the  time-phase  angle  between  the 
voltages  of  a  polyphase  alternator  is  equal  to  360  degrees  divided 
by  the  number  of  phases.  This  statement  does  not  hold  for  a 
so-called  two-phase  alternator,  which  has  the  equivalent  of  two 
armature  windings  ninety  degrees  apart  instead  of  four  as  in  the 
case  of  a  four-phase  machine.  The  only  difference  between  two- 
phase  and  four-phase  alternators  is  in  the  manner  in  which  their 
armature  windings  are  interconnected.  A  two-phase  system  is 
really  half  of  a  four-phase  system. 

The  arrangement  of  a  two-pole,  two-phase  alternator  with  a 
revolving  armature  is  shown  in  Fig.  69.  The  slip  rings  for  the 
collection  of  the  two-phase  currents  are  omitted  to  avoid 
confusion. 

245 


246  PRINCIPLES  OF  ALTERNATING  CURRENTS 

A  polyphase  alternator  always  has  as  many  independent 
armature  windings  as  there  are  phases.  These  are  displaced 
from  one  another  on  the  armature  by  360  electrical  degrees 
divided  by  the  number  of  phases.  In  space  degrees  this  displace- 
ment would  correspond  to  360  degrees  divided  by  the  number  of 

phases  and  by  the  number  of  pairs 
of  poles. 

With  very  few  exceptions,  all 
modern  alternators  are  three-phase. 
Very  few  two-phase  or  four-phase 
alternators  are  built,  except  for  use 
in  existing  two-phase  or  four-phase 
plants.  All  power  transmission  is 
three-phase.  With  fixed  voltage 

between  any  two  conductors  of  a  transmission  line  and  fixed 
transmission  loss,  a  greater  amount  of  power  can  be  transmitted 
a  given  distance  three-phase  than  with  any  other  number  of  phases. 
With  fixed  voltage  to  neutral,  the  amount  of  copper  required  for  a 
transmission  line  is  independent  of  the  number  of  phases.  The 
limiting  condition,  however,  is  not  the  voltage  to  neutral  but  the 
maximum  voltage  between  any  pair  of  conductors.  The  cost  of 
the  greater  number  of  insulators  required  for  a  four-phase  trans- 
mission line  over  the  number  necessary  for  a  three-phase  trans- 
mission line  would  alone  be  a  serious  item. 

All  commercial  alternators  have  stationary  armatures  and 
rotating  fields.  Such  an  arrangement  requires  no  slip  rings 
for  the  collection  of  the  armature  current.  Two  slip-rings  are 
required  for  the  field  winding,  but  these  have  to  carry  only 
the  low-voltage  direct-current  necessary  for  excitation.  The 
armature  of  an  alternator  consists  of  a  laminated  steel  core 
slotted  on  its  inside  for  the  armature  winding.  The  poles  revolve 
inside  the  armature  core.  This  arrangement  puts  the  more 
complicated  winding,  and  the  one  which  is  subjected  to  high 
voltages,  on  the  stationary  part  of  the  machine  where  it  may  be 
easily  insulated.  It  also  relieves  the  armature  winding  from  all 
stresses  except  those  due  to  the  load. 

A  four-pole,  three-phase  alternator  with  a  revolving  field 
and  a  stationary  armature  is  shown  in  Fig.  70.  1-1',  2-2'  and 
3-3'  are  the  coils  for  the  three  phases.  Since  there  are  four 


POLYPHASE  CURRENTS 


247 


poles,  there  are  two  groups  of  coils  for  each  phase.  The  two 
groups  of  coils  for  any  phase  are  360  electrical  degrees  apart. 
The  voltages  induced  in  them,  therefore,  are  in  time-phase. 
Since  the  voltages  induced  in  the  two  coils  for  any  phase  are 
in  time-phase,  these  coils  may  be  connected  either  in  series  or 
in  parallel.  They  would, 
however,  be  connected  in 
series  in  most  cases. 

All  alternators,  except  those 
built  to  be  driven  by  high- 
speed steam  turbines,  have 
more  than  two  poles.  Large 
low-speed  alternators,  such 
as  those  designed  to  be 
driven  by  low-head  water 
wheels,  may  have  as  many 
as  forty  or  fifty  poles.  The 
frequency  of  an  alternator  is 
independent  of  the  number  of 
phases  for  which  it  is  wound. 
It  depends  only  on  the  number 
Frequency  is  always  given  by 


FIG.  70. 


of    poles    and   the   speed. 


rev.  per  mm 


-'  X  no.  pairs  of  poles  =  cycles  per  second.     (1) 


60 

The  armature  winding  of  each  phase  of  multipolar  alternators 
generally  consists  of  as  many  identical  groups  of  windings  as  there 
are  pairs  of  poles.  '  These  are  displaced  360  electrical  degrees  from 
one  another  on  the  armature  and  therefore  have  voltages  induced 
in  them  which  are  in  time-phase.  Since  the  voltages  are  in  time- 
phase,  the  groups  of  windings  for  any  phase  may  be  connected 
either  in  series  or  in  parallel.  Series  connection,  however,  is  the 
more  common,  as  high  voltage  is  usually  desired.  In  general,  the 
windings  of  a  polyphase  alternator  are  spread  out  so  as  to  cover 
the  entire  armature  periphery.  This  distributes  the  heating  due 
to  the  armature  copper  loss  and  also  improves  the  wave  form.  It 
decreases  the  resultant  voltage  obtained  from  a  given  number 
of  armature  turns  and  a  fixed  field  flux,  but  this  diminution  is 
small  in  the  case  of  polyphase  alternators,  only  four  or  five  per 


248  PRINCIPLES  OF  ALTERNATING  CURRENTS 

cent,  and  is  more  than    compensated  for  by  the  better  distri- 
bution of  the  armature  copper  loss  and  the  better  wave  form. 

The  chief  advantages  of  three-phase  alternators  over  single- 
phase  alternators  are: 

1.  They  give  about  fifty  per  cent  greater  output  for  a  given 
amount  of  material.     For  a  given  capacity,  therefore,  they  are 
cheaper. 

2.  They  are  more  satisfactory  for  the  operation  of  motors, 
since  polyphase  motors  have  better  operating  characteristics 
than   single-phase   motors.     Polyphase   motors   give   the   same 
increase  in  output  for  a  given  amount  of  material  as  polyphase 
alternators. 

3.  They  are  more  efficient. 

4.  They  are  more  satisfactory  for  power  transmission,  since 
with  a  fixed  voltage  between  conductors,  fixed  amount  of  power 
transmitted   a  given   distance   and   a   fixed   transmission   loss, 
twenty-five   per   cent   less    copper   is  required  for  three-phase 
transmission  than  for  single-phase  transmission.     The  copper 
efficiencies  of  single-phase  and  of  four-phase  transmission  are  the 
same. 

Double  Subscript  Notation  for  Polyphase  Circuits. — In  work- 
ing with  current  and  voltage  relations  in  the  simplest  polyphase 
circuits,  the  particular  notation  employed  is  not  important.  In 
the  more  complex  cases,  however,  any  lack  of  definiteness  in  the 
notation  or  its  interpretation  is  almost  sure  to  result  in  confusion, 
if  not  in  serious  error.  A  simple  and  very  satisfactory  notation 
is  based  on  lettering  every  junction  and  terminal  point  of  dia- 
grams of  connections  and  on  the  use  of  two  subscripts  with 
every  vector  representing  current  or  voltage. 

The  vector  diagram  must  be  distinguished  from  the  diagram 
of  connections,  although  in  certain  cases  there  may  be  some 
similarity  between  them.  The  subscripts,  taken  from  the  dia- 
gram of  connections,  indicate  that  the  positive  direction  of  the 
current  is  from  the  first  to  the  second  and  that  the  positive  direc- 
tion of  the  elecromotive  force  is  also  from  the  first  to  the  second. 
The  current  /o&,  according  to  this  notation,  is  the  current  whose 
positive  direction  is  from  a  to  b  in  the  branch  ab  of  the  circuit, 
and  Eab  is  the  electromotive  force  which  produces  this  current. 
Also  7ba  is  the  current  whose  positive  direction  is  from  6  to  a 


POLYPHASE  CURRENTS  249 

and  it  is  produced  by  the  electromotive  force  Eba.  Again,  lab  is 
equal  to  —  /&<.,  and  7  a*  and  ha  differ  in  phase  by  180  degrees. 
Eab  and  Eba  also  differ  by  180  degrees.  If  the  rotating  vector 
marked  7,  Fig.  21,  page  66,  is  the  current  lab,  it  is  positive  during 
the  time  its  projection  on  the  vertical  axis  is  positive,  that  is, 
when  the  revolving  vector  7  lies  in  the  first  or  second  quadrants. 
During  this  time  the  current  actually  flows  from  a  to  6  in  the 
circuit  ab.  While  the  current  vector  7  lies  in  the  third  or  fourth 
quadrants,  the  current  actually  flows  from  6  to  a.  If  the  vector 
E,  Fig.  21,  page  66,  represents  the  rise  in  potential  Eab,  there  is 
an  actual  rise  in  potential  in  the  direction  a  to  &  during  the  time 
the  projection  of  the  vector  E  on  the  vertical  axis  is  positive. 
There  is  a  fall  of  potential  in  the  direction  ab  when  the  projection 
of  the  vector  E  on  the  vertical  axis  is  negative.  (See  pages  64 
and  65.) 

If  Eab  represents  a  rise  in  potential,  then,  on  the  average, 
power  will  be  delivered  during  each  cycle  when  Eab  and  7o& 
have  components  in  phase,  and  power  will  be  absorbed  when 
they  have  components  in  opposition.  Power  delivered  will  be 
positive  and  power  absorbed  will  be  negative.  If  Eab  represents 
a  fall  in  potential  —  Eab  =  Eba  will  be  the  corresponding  potential 
rise.  In  this  case  power  will  be  absorbed  when  Eab  and  7o& 
have  components  in  phase  and  it  will  be  delivered  when  they 
have  components  in  opposition.  If  the  circuit  ab  contains  pure 
resistance,  the  current  Iab  will  be  in  phase  with  the  voltage  drop 
produced  by  the  current  in  the  resistance.  It  will  be  in  phase 
with  Eab  when  the  vector  Eab  represents  a  voltage  drop.  It  will 
be  in  phase  with  —  E,j>  when  the  vector  Eab  represents  a  voltage 
rise.  If  the  circuit  ab  contains  inductive  reactance  in  addition 
to  resistance,  lab  will  lag  the  voltage  drop  by  an  angle  whose 
cosine  is  the  ratio  of  the  resistance  to  the  impedance,  that  is  by 
the  angle 

0  =  cos-1  -T=  (2) 

V  r2  +  x2 

where  r  and  x  are  the  resistance  and  reactance  respectively  of  the 
circuit  ab. 

The  current  7o&  must  always  be  used  with  the  voltage  Eab. 
The  power  in  the  circuit  is  Eablab  cos  0  and  never  Eablba  cos  0. 
If  Eab  is  a  voltage  rise,  Eablab  cos  0  is  power  generated.  If 


250  PRINCIPLES  OF  ALTERNATING  CURRENTS 


is  a  voltage  drop,  EaJ.ab  cos  0  is  power  absorbed.  If  Eablab  cos  6, 
representing  power  generated,  is  negative,  it  actually  represents 
power  absorbed,  since  negative  power  generated  is  power  absorbed. 
HEablab  cos  '0,  representing  power  absorbed,  is  negative,  it  actually 
represents  power  delivered  or  generated. 

The  voltage  drop  Eab  due  to  a  current  in  an  impedance  is 


Eab    =    lab  V^2  +  X2 

It  is  never  Eab  =  Iba\/r2  -+-  x2. 

It  is  often  necessary  to  use  double  subscripts  with  resistance, 
reactance  and  impedance  and  also  with  conductance,  susceptance 
and  admittance,  but  in  such  cases  the  subscripts  have  no  other 
significance  than  to  indicate  between  what  two  points  of  a  circuit 
the  quantities  mentioned  are  measured.  The  order  of  the  sub- 
scripts can  mean  nothing,  since  resistance,  reactance,  impedance, 
etc.  are  not  vectors. 

When  in  specific  problems  exact  numerical  results  are 
required,  an  analytical  solution  by  the  method  of  complex 
quantities  is  necessary.  However,  experience  has  shown  con- 
clusively that,  even  though  a  particular  problem  is  to  be  solved 
analytically,  an  approximate  vector  diagram  should  always  be 
drawn  as  it  facilitates  the  correct  interpretation  of  the  work 
and  serves  as  a  check  against  errors.  It  is  usually  advisable 
to  draw  all  polyphase  vectors  radially  from  a  common  center. 
When  vectors  are  drawn  radially  from  a  common  center,  there 
can  be  no  question  as  to  their  phase  relations,  and  if  each  vector 
is  lettered  with  two  subscripts,  ihere  is  no  excuse  for  mistaking 
the  angle  between  any  two  vectors,  such  as  Eab  and  7Cd,  for  the 
angle  between  Eab  and  Idc-  Neither  is  there  any  excuse  for 
mistaking  the  angle  between  Eab  and  7ba  for  the  angle  between 
Eab  and  7a&.  The  one  angle  is  the  supplement  of  the  other. 

Example  Illustrating  the  Use  of  Double -subscript  Notation.— 
A  certain  three-phase  generator  has  one  end  of  each  of  its  three 
armature  windings  connected  together  in  such  a  manner  that  the 
voltages  generated  in  the  windings,  when  considered  in  a 
direction  from  common  junction  to  free  terminal,  are  120 
degrees  apart  in  time-phase.  The  three  voltages  will  ""Be  as- 
sumed to  be  equal  in  magnitude.  The  diagram  of  connections 
is  shown  in  Fig.  7 la. 


POLYPHASE  CURRENTS 


251 


Let  inductive  loads,  n  +  jxi  and  r2  -f  jx2,  be  apph'ed  between 
the  terminals  ab  and  ac  respectively,  as  shown  in  Fig.  7  la. 
Call  V,  with  proper  subscripts,  the  voltage  rise  across  the  termi- 
nals of  the  generator  windings.  Let  the  internal  impedance  of 
each  winding  be  r  +  jx.  The  internal  or  induced  voltage  rises 
in  the  three  phases  are  desired. 


(a) 


Ic'a'-I 


Ij.r  +  ix 


FIG.  71. 


Referring  to  the  vector  diagram  shown  in  Fig.  716,  Voa,  V0b 
and  Voc  are  the  voltage  rises  across  the  terminals  of  phases  oa,  ob 
and  oc  of  the  generator.  For  simplicity,  these  voltages  are 
assumed  to  be  equal  in  magnitude  and  120  degrees  apart  in 
time-phase.  Actually,  the  voltage  rises  E^,  E0b  and  Eoc  induced 
in  the  windings  would  be  equal  in  magnitude  and  120  degrees 
apart  in  time-phase  and  the  terminal  voltages  would  differ 
slightly  in  magnitude  and  from  the  120  degree  phase  relationship, 
due  to  the  impedance  drops  in  the  armature  windings  of  the 
alternator.  However,  for  the  illustration  of  the  use  of  double- 
subscript  notation,  the  assumption  of  equal  terminal  voltages, 
120  degrees  apart  in  time-phase,  is  made,  since  it  permits  a  defi- 
nite solution  of  the  problem  without  the  use  of  Kirchhoff  s  laws 
and  therefore  better  suits  the  present  purpose. 

The  voltage  Vab  between  any  pair  of  terminals  such  as  a  and  b  is 
equal  to  the  vector  sum  of  the  voltages  generated  in  the  two 
phases  connected  between  those  terminals,  considered  in  the 
direction  ab.  The  voltage  rise  Vab  is  therefore  equal  to  Vao  + 


252  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Vob  =  —Voa  +  Vob-  It  is  equal  to  the  vector  difference  between 
the  voltage  rises  V0b  and  Voa-  The  voltage  rises  between  the 
terminals  ab,  be  and  ca  are 

Vab    =    V_a'b'    =    -Voa  +  Vob  (3) 

Vbc   =    Vb'c'    =    -Vob   +  VoC  (4) 

Vca   =    FCV    =    -Foe   +  Voa  (5) 

The  corresponding  voltage  drops  are 

-V_ab    =     -Va'b>     =    Voa    ~Vob  (6) 

-Vbc    =     -Vb'c'     =   'Vob    -Voc  (7) 

~Vca    =     -Vc'a'     =    Voc    -Voa  (8) 

The  voltage  drops   —  Vab  and   —  Vca  are  shown  in  Fig.  716. 

The  current  Ja>v  in  the  impedance  Zi,  connected  between  the 
terminals  'a'  and  b',  is  equal  to  the  voltage  drop  —  Vab  =  —Va>b' 
across  a'b'  divided  by  the  impedance  z\  =  r\  +  jx\.  This  current 

lags  the  voltage  drop  —  Va>v  by  the  angle  0i  =  tan"1  — •  Simi- 
larly, the  current  7cv,  in  the  impedance  z2  connected  between  the 
terminals  c'  and  a' ,  is  equal  to  the  voltage  drop  —  Vca  =  —  Vcv 
across  c'a'  divided  by  the  impedance  22  =  r2  +  jx^.  This  cur- 
rent lags  the  voltage  drop  —  Vc>a>  by  the  angle  02  =  tan"1  — • 

The  current  in  the  phase  oa  of  the  generator  must  be  equal  to 
the  current  in  the  line  connected  to  that  phase.  This  line  current 
must  be  equal  to  the  vector  sum  of  the  currents  in  the  two 
impedances  connected  to  this  line.  Therefore 

loa    =   laa'    =   la'b'      I     la'c' 

=   la'b'    ~  /c'a'  (9) 

The  currents  carried  by  the  other  two  phases  of  the  generator, 
i.e.,  phases  ob  and  oc,  are 

Lb    =  hb'    =   /6V  (10) 

loc    =   I  cc'    =   I  c'a'  (11) 

The  voltage  rise  induced  in  any  armature  winding,  i.e.,  in 
any  armature  phase,  is  equal  to  the  vector  sum  of  the  voltage 
rise  across  its  terminals  and  the  impedance  drop  produced  in  it 


POLYPHASE  CURRENTS  253 

by  the  current.     If  E^  E0b  and  Eoc  are  respectively  the  voltage 
rises  induced  in  the  phases  oa,  ob  and  oc,  then 

Eob  =  V0b  +  Iob(r  +  jx)  (13) 

Eoc  =  Voc+Ioc(r+jx)  (14) 

In  order  to  illustrate  the  analytical  solution  ^f  this  problem, 
assume  that  the  three  terminal  voltages  Voa,  V0b  and  Voc  are 
each  equal  to  200  volts.  They  are  also  assumed  to  be  120 
degrees  apart  in  time-phase.  Let  the  two  impedances  which 
serve  as  loads  be 

21  =  n  +  jxi  =  20  +  j5  ohms  (15) 

22  =  7*2  +  jx2  =  15  +  J15  ohms  (16) 
Let  the  impedance  of  each  phase  of  the  generator  be 

z  =  r  +  jx  =  0.1  +  jl.O  ohm  (17) 

Take  the  terminal  voltage  rise  Voa  along  the  axis  of  imagi- 
naries,  to  correspond  to  the  vector  diagram  of  Fig.  716.  Then 

Voa  =  200(cos  90°  +  j  sin  90°) 

=  200(0  +  jl) 

=  0  -f  J200 
Vob=  200(cos  30°  -  j  sin  30°) 

=  200(0.866  -  jO.500) 

=  173.2  -  jlOO 
Voc=  200(cos  150°  -  j  sin  150°) 

=  200( -0.866  -jO.500) 

=  -173.2  -jlOO 

From  equations  (6)  and  (8) ,  the  voltage  drops,  —  Vab  and  —  F^,, 
across  the  impedances  z\  and  32  are 

-Vab    =    -Va'b'    =    Voa    ~   V ' ob     =    (0  +  J200)    -    (173.2    -  J100) 

=  -173.2  +  J300  volts 

-V^  =  -V  C'a'=  V0c  -Voa   =  (-173.2  -  jlOO)  -  (0 

=  -173.2  -J300  volts 

la'b'  —  ~ 


- 173.2 -f  J300 
20  +  J5 


254  PRINCIPLES  OF  ALTERNATING  CURRENTS 

- 173.2  -f  J300        20  -  j5 
20  +  j5  20  -  j5 

=  —4.62  +  J16.14  amperes. 


-173.2  -  J300 

15  +  J15 

-173.2  -  J300      15  -  j!5 
15  +  j!5  15  -  j!5 

=  —  15.78  —  J4.23  amperes. 

J-  oa   =   J-a'b'          'c'a' 

=  (-4.62  +  J16.14)  -  (-15.78  -  J4.23) 
=  11.16  +  J20.37  amperes. 

lob    =    Ib'a'    —     ~Ia'b' 

=  4.62  —  J16.14  amperes. 

J-oc          J-  c'a' 

=  -15.78  -  J4.23  amperes. 

Eoa   =    Voa  +  Ioa(r   +  jx) 

=  (0  +  J200)  +  (11.16  +  j20.37)(0.1  +  jl.O) 
=  -19.25  +  J213.2 


Eoa  =       (-19-25)2 

=  214.1  volts. 
Eob  =  Vob  +  Iob(r  +  jx) 

=  (173.2  -  J100)  +  (4.62  -  J16.14)(0.1  +J1.0) 

=  189.8  -  J97.0 
Eob  ==  \/(189.8)2  +  (-97^2 

=  213.2  volts. 

Eoc    =    Voc   +  Ioc(r   +  JX) 

=  (-173.2  -  jlOO)  +  (-15.78  -j4.23)(0.1  +  jl.O) 
=  -175.9  -  J115.2 


Eoc  =  V(-175.9)2  +  (- 
=  210.2  volts. 

Call  the  angles  made  by  the  vectors  Eoa,  E0b  and  Eoc  with  the 
axis  of  reals  Boaj  Oob  and  doc.     Then 


POLYPHASE  CURRENTS 


255 


Since  the  real  part  of  the  voltage  Eoa  is  negative  and  the 
imaginary  part  is  positive,  the  angle  £«,  must  lie  in  the  second 
quadrant.  Therefore 

Boa  =  95.2  degrees. 

-97.0 

tan  B0b  = 


189.8 


-0.511 


Since  the  real  part  of  the  voltage  Eob  is  positive  and  the  im- 
aginary part  is  negative,  the  angle  Bob  must  lie  in  the  fourth 
quadrant.  Therefore 

B0b  =  —27.2  degrees. 
-115.2 


tan  doc  = 


-175.9 


=  0.655 


Since  the  real  part  of  the  voltage  Eoc  is  negative  and  the 
imaginary  part  is  also  negative,  the  angle  Boc  must  lie  in  the 
third  quadrant.  Therefore 


BOC  =  —146.8  degrees. 

i 

The  polar  expressions  for  the  three  voltages 
Eoc  are 

EM  =  214.1 95°2 


E0b  and 


213.227?2 


Eoc  =  210.2  1468 

Wye  and  Delta  Connections  for  Three-phase  Generators  and 
for  Three-phase  Circuits.  —  The  windings  of  a  three-phase  alter- 
nator may  be  represented  dia- 
grammatically  by  three  windings 
displaced  120  degrees  from  one 
another,  as  shown  in  Fig.  72. 

If  the  windings  are  assumed  to 
rotate  in  a  counter-clockwise  di- 
rection, the  voltages  generated  in 
them  will  be  equal  and  will  differ 
120  degrees  in  time-phase.  The 
voltage  generated  in  the  phase 

marked  01  will  lead  the  voltage  generated  in  phase  02  by  120 
degrees.  It  will  lead  the  voltage  generated  in  phase  03  by 
240  degrees.  Three-phase  voltages  will  be  generated  in  the 


FIG.  72. 


256 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


windings.  These  may  be  represented  by  three  vectors,  #0i, 
#02  and  Eos,  which  are  equal  in  magnitude  and  120  degrees  apart 
in  time-phase.  These  are  shown  in  Fig.  73.  The  order  of  the 
subscripts  used  with  the  vectors  indi- 
cates the  direction  in  which  the  voltages 
are  considered  with  respect  to  the 
windings.  All  voltages  must  be  con- 
sidered in  the  same  direction  around 
the  armature  if  they  are  to  differ  by  120 
degrees  in  time-phase. 

The  voltage  vector  Eoi  is  taken  along 
the  axis  of  reals.     The  symbolic  expres- 
sions for  the  three  equal  voltages  are 
#01  =  (1  -  J0)#  (18) 


FIG.  73. 


-<-$-*£>• 


J^  rn    —    (  —  ~ 


(19) 


(20) 


where  E  is  the  magnitude  of  the  voltage  generated  in  any  one 
phase.     From  equations    (18),  (19)  and  (20) 


+  #02    +#03    =0 


(21) 


Since  the  vector  sum  of  the  voltages  generated  in  the  three 
windings  is  equal  to  zero  when  considered  in  the  same  direction 
around  the  armature,  the  terminals  of  windings  01,  02  and  03 
may  be  connected  to  form  a  closed  circuit,  as  shown  in  the  left- 
hand  half  of  Fig.  74,  and  no  current  will  flow  since  the  resultant 
voltage  acting  in  the  closed  circuit  formed  by  the  armature 
windings  is  zero.  The  terminals  of  the  alternator  will  be  taken 
from  the  junction  points  between  the  phases,  i.e.,  from  the  points 
1,  2  and  3.  The  windings  connected  in  this  manner  form  a 
closed  delta.  This  connection  is  known  as  the  delta  connection. 

Instead  of  connecting  the  phases  in  delta,  they  may  be  con- 
nected to  form  a  wye  by  joining  the  corresponding  terminals  of 
the  windings  of  the  three  phases.  Either  the  terminals  marked  0 
or  those  marked  1,  2  and  3  may  be  joined  for  wye  connection. 
Delta  and  wye  connections,  usually  written  A  and  Y,  are  illus- 
trated in  Fig.  74.  The  field  poles  are  omitted  in  the  figure. 


POLYPHASE  CURRENTS  257 

A  A-connected  alternator  can  have  only  three  terminals.  A 
Y-connected  alternator  need  have  only  three,  but  in  some  cases  a 
fourth  is  brought  out  from  the  common  junction  between  the 
windings  or  neutral  point,  as  this  junction  is  called,  to  permit 
grounding  the  alternator.  When  loads  are  connected  in  wye  the 
neutral  connection  is  usually  employed.  When  the  neutral 
point  is  available,  the  load  may  be  applied  between  the  three 
pairs  of  line  terminals,  i.e.,  between  1  and  2,  between  2  and  3 
and  between  3  and  1,  or  it  may  be  applied  between  the  terminals 
1,  2  and  3  and  the  neutral  point.  In  the  first  case  the  load  is 


A  -Connection  3f-Connection 

FlG.    74. 

A-connected.  In  the  second  case  it  is  Y-connected.  When  the 
windings  of  an  alternator  are  connected  in  delta,  the  load  is 
usually  connected  in  delta,  although  it  may  be  connected  in 
wye;  but  when  the  load  is  connected  in  wye  there  can  be  no 
connection  between  the  neutral  of  the  load  and  the  generator 
since  no  neutral  connection  for  the  alternator  exists.  Although 
both  Y-  and  A-connected  loads  are  used,  A-connection  is  the  more 
common.  Y-connection  is  the  more  common  for  alternators. 
Y-connected  alternators  are  somewhat  better  than  those  con- 
nected in  delta  for  several  reasons,  the  most  important  of  which 
are:  there  can  be  no  short-circuit  current  in  their  armatures  due 
to  harmonics;  for  a  fixed  terminal  voltage  and  output,  the  ratio 
of  the  amount  of  insulation  to  copper  required  in  their  armature 
windings  is  somewhat  less  than  for  A-connection;  they  permit 
grounding  the  neutral  point  of  the  system. 

If  an  alternator  has  more  than  two  phases,  there  are  always 
two  ways  in  which  its  armature  windings  may  be  connected. 
These  are  known  as  the  mesh  and  the  star  connections  and 

17 


258  PRINCIPLES  OF  ALTERNATING  CURRENTS 

correspond  to  the  delta  and  wye  connections  for  a  three-phase 
alternator.  Alternators  with  more  than  four  phases  are  not 
used.  They  would  possess  no  advantage  over  three-phase  or 
four-phase  alternators  and  would  have  marked  disadvantages 
from  the  standpoint  of  power  transmission.  They  would  also 
be  more  complicated.  Very  few  four-phase  alternators  are 
built,  except  for  special  purposes  such  as  for  use  in  existing  four- 
phase  plants.  From  the  standpoint  of  power  transmission, 
three-phase  is  superior  to  all  others.  It  requires  only  three  line 
conductors,  as  against  four  for  four-phase  and  six  for  six-phase, 
and  for  a  fixed  amount  of  power  transmitted  a  fixed  distance 
with  a  fixed  line  loss  and  fixed  voltage  between  any  two  con- 
ductors, (not  necessarily  adjacent),  it  requires  only  three-quarters 
as  much  copper  as  single-phase,  four-phase  or  six-phase.  From 
the  standpoint  of  power  transmission  alone,  four-phase  trans- 
mission possesses  no  advantage  over  single-phase  transmission. 
All  power  transmission  lines  are  three-phase. 

Relative  Magnitudes  and  Phase  Relations  of  Line  and  Phase 
Currents  and  of  Line  and  Phase  Voltages  for  a  Balanced  Three- 
phase  System  Having  Sinusoidal  Current  and  Voltage  Waves. — 
By  a  balanced  system  is  meant  one  in  which  the  voltages  in  all 
phases  are  equal  in  magnitude  and  differ  in  phase  by  equal  angles. 
The  currents  must  also  be  equal  in  magnitude  and  they  must 
also  differ  in  phase  by  equal  angles.  For  a  balanced  system  the 
phase  angles  between  the  voltages  and  also  between  the  currents 

360 
are  equal  to degrees,  where  n  is  the  number  of  phases.     A 

n 

balanced  load  is  one  in  which  the  loads  connected  across  all 
phases  are  identical. 

For  Y-connection,  there  can  obviously  be  no  difference  between 
the  current  in  any  phase  and  the  current  in  the  line  to  which  the 
phase  is  connected.  Line  and  phase  currents  are  the  same  for  Y- 
connection.  Line  and  phase  voltages,  however,  are  not  the 
same.  The  voltage  between  any  two  terminals,  i.e.j  the  line 
voltage,  is  the  vector  difference  of  the  voltages  in  the  phases 
connected  between  the  two  terminals  considered.  The  line 
voltage  is  neither  equal  in  magnitude  to  the  phase  voltage  nor 
is  the  line  voltage  in  phase  with  the  phase  voltage.  Refer  to 


POLYPHASE  CURRENTS 


259 


Line  current  In 
Line  current  722 
Line  current  /33 


FIG.  75. 


Fig.  75.     For  Y-connection,  the  same  diagram  may  be  used  for 
either  a  diagram  of  connections  or  a  vector  diagram.     Fig.  75 
will  serve  for  both. 
Obviously 

=  phase  current  loi. 
=  phase  current  I02. 
=  phase  current  /os. 

The  line  voltage  Viz  is  not  equal  in 
magnitude  to  either  Voi  or  Voz  or  in  3 
phase  with  either. 

Viz  =  Vio  +  Fo2_ 
=    —  Voi  +  Voz 

Viz  =  2V  cos  30° 

=  \/37  in  magnitude        (22) 

where  V  is  the  magnitude  of  the  phase  voltage. 

Viz  is  equal  in  magnitude  to  the  phase  voltage  multiplied  by 
the  square  root  of  three.  It  differs  in  phase  from  the  phase 
voltage  Voz  by  30  degrees  and  from  the  phase  voltage  Voi  by  150 
degrees. 

For  a  balanced,  three-phase,  Y-connected  circuit,  line  voltage  is 
equal  in  magnitude  to  phase  voltage  multiplied  by  the  square 
root  of  three.  Line  voltage  differs  in  phase  from  the  voltages 
in  the  phases  connected  between  the  lines  considered  by  30  or 
150  degrees,  according  to  which  of  the  two  phase  voltages  is 
considered. 

Viz  lags  Voz  by  30  degrees  and  lags  Voi  by  150  degrees.  (23) 
Vzs  lags  Vos  by  30  degrees  and  lags  Voz  by  150  degrees.  (24) 
Vzi  lags  Voi  by  30  degrees  and  lags  Voz  by  150  degrees.  (25) 

If  the  phase  rotation  is  opposite  to  that  shown  in  Fig.  74,  i.e., 
if  Voz  leads  Voi  by  120  degrees  and  Vos  leads  Voi  by  240  de- 
grees, Viz  would  lead  Voz  by  30  degrees  instead  of  lagging  it  by 
30  degrees,  Vza  would  lead  Vw  by  30  degrees  instead  of  lagging 
it  by  30  degrees  and  Vai  would  lead  Voi  by  30  degrees  instead  of 
lagging  it  by  30  degrees. 

The  relations  existing  between  phase  and  line  currents  of 
a  balanced  A-connected  system  are  similar  to  those  existing 


260 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


between  phase  and  line  voltages  of  a .  balanced  Y-connected 
system.  Let  the  left-  and  right-hand  diagrams,  Fig.  76,  be, 
respectively,  a  diagram  of  connections  and  a  vector  diagram  of 

currents  in  the  branches, 
/u/  i.e.,  in  phases,  12,  23  and 

31  of  a  A-connected  arma- 
ture or  A-connected  load. 
The  line  current  7ir  in 
line  11',  which  is  con- 
nected to  the  common 
junction  point  of  phases 
12  and  31,  is  equal  to  the 
vector  sum  of  the  currents 
in  these  phases,  both  con- 
6  sidered  in  a  direction  to- 

wards the  junction  point, 

1.  This  follows  from  the  fact  that  the  vector  sum  of  the  currents 
at  any  junction  point  must  be  equal  to  zero. 


For  a  balanced  load 


/!!'    =    /21    +  /31 

=     -7l2    +   hi 

Iu>.  =  21  cos  30° 

=  \/37  in  magnitude, 


(20) 


where  /  is  the  magnitude  of  the  phase  current. 

The  line  current  7ir  is  equal  in  magnitude  to  the  phase  current 
multiplied  by  the  square  root  of  three.  It  differs  in  phase 
from  the  phase  current  /si  by  30  degrees  and  from  the  phase  cur- 
rent 7i2  by  150  degrees. 

For  a  balanced  three-phase  A-connected  circuit,  line  current 
is  equal  to  phase  current  multiplied  by  the  square  root  of  three. 
Line  current  differs  in  phase  from  the  currents  in  the  phases  to 
which  the  line  considered  is  connected,  by  either  30  or  150  de- 
grees according  to  which  of  the  two  phases  is  considered. 

In'  leads  7si  by  30  degrees  and  leads  In  by  150  degrees.  (27) 
/22'  leads  7 12  by  30  degrees  and  leads  Jzs  by  150  degrees.  (28) 
733'  leads  72a  by  30  degrees  and  leads  73i  by  150  degrees.  (29) 


POLYPHASE  CURRENTS  261 

If  the  phase  rotation  were  opposite  to  that  shown  in  Fig.  76, 
Iu>  would  lag  73i  by  30  degrees,  722'  would  lag  7i2  by  30  degrees 
and  /sa'  would  lag  7  23  by  30  degrees. 

In  a  A-connected  system,  the  voltage  between  any  pair  of 
lines,  such  as  1  and  2,  is  obviously  equal  to  and  in  phase  with  the 
voltage  in  the  phase  connected  between  the  lines  considered. 
It  is  also  equal  to  the  vector  sum  of  the  voltages  in  the  other  two 
phases  when  these  are  considered  in  the  proper  direction. 

Fi2  =  Fi3  +  F32  (30) 

F23    =    721   +   Fl3  (31) 

Fn  =  F32  +  F»i  (32) 

For  a  A-connected  system,  line  and  phase  voltages  are  equal. 
If  a  single-phase  load  is  applied  between  any  pair  of  terminals 
of  a  A-connected  alternator,  the  current  will  divide  between  the 
two  parallel  paths  formed  by  the  armature  windings  inversely  as 
their  impedances.  The  impedance  of  the  windings  is  the  same 
for  each  phase  of  an  alternator.  Since  one  path  consists  of  a 
single  phase  and  the  other  path  consists  of  two  phases  in  series, 
the  impedances  of  the  two  paths  between  which  the  current  divides 
will  be  in  the  ratio  of  one  to  two.  Therefore,  the  phase  across 
which  the  load  is  applied  will  carry  two-thirds  of  the  current. 
The  other  two  phases  in  series  will  carry  the  remaining  third. 
The  currents  in  the  two  branches  will  be  in  phase,  since  the  ratio 
of  the  resistance  to  the  reactance  is  the  same  in  each  branch. 

When  a  balanced  three-phase  load  is  connected  across  the 
terminals  of  an  alternator  which  has  balanced  voltages,  the  three 
line  currents  will  be  equal  in  magnitude  and  will  differ  by  120 
degrees  in  phase.  The  three  phase-currents  in  the  windings  of  the 
alternator  will  also  be  equal  in  magnitude  and  will  differ  by  120 
degrees  in  phase.  The  phase  currents  will  be  the  same  as  the  line 
currents  for  a  Y-connected  alternator  and  will  be  equal  to  the  line 
currents  divided  by  the  square  root  of  three  for  a  A-connected 
alternator.  For  the  A-connected  alternator,  there  will  be  a  shift 
in  phase  between  line  and  phase  currents,  as  is  shown  by  equations 
(27),  (2'8)  and  (29).  The  phase  angle  between  the  phase  currents 
and  phase  voltages  will  be  the  same  for  the  alternator  as  for  the 
load.  These  statements  are  entirely  independent  of  whether  the 
load  and  generator  are  connected  alike,  i.e.,  both  either  in  wye  or 


262  PRINCIPLES  OF  ALTERNATING  CURRENTS 

in  delta,  or  whether  they  are  connected  differently,  i.e.,  one  in 
wye  the  other  in  delta. 

Relative  Magnitudes  and  Phase  Relations  of  Line  and  Phase 
Currents  and  of  Line  and  Phase  Voltages  of  a  Balanced  Four- 
phase  System  Having  Sinusoidal  Current  and  Voltage  Waves. — 
|p  A  four-phase  alternator  has  four  identical 
armature  windings  which  are  displaced 
ninety  electrical  degrees  from  one  another. 
These  windings  may  be  connected  either 
~^  in  star  or  in  mesh,  corresponding  to  the 
wye  and  delta  connections  for  a  three-phase 
alternator.  The  voltages  induced  in  the 
^  windings  of  a  four-phase  alternator  are 

ninety  degrees  apart  in  time-phase  and  may 
therefore  be  represented  by  four  equal  vectors  displaced  ninety 
degrees  from  one  another  as  shown  in  Fig.  77. 
The  vector  expressions  for  the  four  vectors  are 

7oi  =  7(0 +jl) 

7o2  =  v(i  +  jo) 

703  =  7(0  -  jl) 

704  =  7(-l  +JO) 

where  7  is  the  magnitude  of  the  phase  voltage. 
Since 

7oi  +  702  +  703  +  704  =  0,  (33) 

no  current  will  flow  if  the  four  armature  windings  are  connected 
in  order  around  the  armature  to  form  a  closed  circuit,  i.e.,  are 
connected  in  mesh,  in  the  same  way  as  the  armature  windings  of 
a  three-phase  generator  are  connected  in  delta.  Instead  of 
connecting  the  armature  windings  in  mesh,  they  may  be  con- 
nected in  star  by  joining  their  corresponding  ends  in  the  same 
way  that  the  windings  of  a  three-phase  alternator  are  joined 
for  wye  connection. 

Mesh  and  star  four-phase  connections  are  shown  in  Fig.  78. 
For  simplicity,  gramme-ring  windings  are  shown,  instead  of 
windings  like  those  actually  used.  The  windings  actually  used 
would  be  similar  to  those  illustrated  in  Fig.  70,  page  247,  for  a 
three-phase  alternator. 

A  four-phase  alternator  must  have  four  terminals.     When 


POLYPHASE  CURRENTS 


263 


connected  in  star,  a  four-phase  alternator  may  have  a  fifth 
terminal  brought  out  from  the  common  junction  between  the 
windings,  i.e.,  from  the  neutral  point. 

In  general,  for  star  connection  it  is  obvious  that  there  can  be 
no  difference  between  the  current  in  any  line  and  the  current  in 
the  phase  to  which  the  line  is  connected.  Line  and  phase  cur- 
rents are  the  same  for  star  connection.  Line  and  phase  voltages 
are  not  the  same  for  star  connection.  The  voltage  between  any 
pair  of  line  terminals,  such  as  1  and  2,  of  a  star-connected  alter- 
nator, i.e.,  the  line  voltage,  is  equal  to  the  vector  sum  of  the 
voltages  in  the  phases  connected  between  the  line  terminals  1 
and  2,  both  voltages  being  considered  in  the  direction  from  1  to  2 
through  the  windings. 


Mesh  Connection 


Star  Connection 


FIG.  78. 


In  general,  for  mesh  connection  the  voltage  between  any  pair  of 
adjacent  line  terminals,  i.e.,  the  line  voltage,  is  equal  to  the  volt- 
age in  the  phase  to  which  the  lines  considered  are  connected. 
Line  and  phase  voltages  are  equal  for  mesh  connection.  Line  and 
phase  currents,  however,  are  not  equal  for  mesh  connection. 
The  current  per  terminal,  i.e.,  the  line  current  of  a  mesh-con- 
nected alternator,  is  equal  to  the  vector  sum  of  the  currents  in 
the  two  phases  to  which  the  line  considered  is  connected,  both 
currents  being  taken  in  the  same  direction  with  respect  to  the 
junction  point  between  the  phases  and  line,  i.e.,  both  being  taken 
either  towards  or  away  from  the  junction  point. 

Refer  to  Figs.  77  and  78.  For  a  four-phase,  star-connected 
alternator, 


Line  current  7n>  =  phase  current  70i. 


Line  current  1W 
Line  current  733' 


=  phase  current  7o2- 
=  phase  current  Jos. 


Line  current  744'  =  phase  current  7  04- 


(34) 
(35) 
(36) 
(37) 


264  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  line  voltage  Viz  is  neither  equal  to  nor  in  phase  with  phase 
voltage  7oi  or  702,  but 


Fi2  =  27  cos  45° 

=  \/2V  in  magnitude, 
where  V  is  the  magnitude  of  the  phase  voltage. 

The  line  voltage  of  a  four-phase,  star-connected  alternator 
is  equal  in  magnitude  to  the  phase  voltage  multiplied  by  the 
square  root  of  two.  It  differs  in  phase  from  the  voltage  in  one 
of  the  phases  between  the  lines  considered  by  45  degrees  and 
from  that  in  the  other  by  135  degrees.  The  line  and  phase 
currents  are  equal. 

For  a  mesh-connected,  four-phase  alternator,  the  current  in 
any  line,  such  as  line  1'  (see  Fig.  78),  is  equal  to  the  vector  sum 
of  the  currents  in  phases  1  and  2  both  taken  towards  the  junction 
point  between  the  line  and  the  phases. 

Ill'    =    loi    ~\~  720 
=    7oi    —    7()2 

where    7oi    and    7o2    are    the    currents    in    phases    01    and  02 
respectively. 

For  a  balanced  load 

Iiv  =  27  cos  45° 

=  -\/2I  in  magnitude,  (38) 

where  7  is  the  magnitude  of  the  phase  currents. 

The  line  current  of  a  four-phase,  mesh-connected  alternator, 
carrying  a  balanced  load,  is  equal  in  magnitude  to  the  phase 
current  multiplied  by  the  square  root  of  two.  There  is  a  phase 
difference  between  the  current  in  any  line  and  in  the  phases  to 
which  it  connects  of  45  or  135  degrees  according  to  which  of  the 
two  phases  is  considered.  The  line  and  phase  voltages  of  a 
mesh-connected,  four-phase  alternator  are  equal. 

Relative  Magnitudes  of  Line  and  Phase  Currents  and  Line  and 
Phase  Voltages  for  Balanced  Star-  and  Mesh-connected  N  -phase 
Systems  Having  Sinusoidal  Current  and  Voltage  Waves.  —  The 
line  voltages  of  any  balanced  mesh-connected  system  are  always 
equal  to  the  phase  voltages.  The  line  and  phase  currents  of  any 
balanced  star-connected  system  are  always  equal.  These 


POLYPHASE  CURRENTS  265 

relations,  which  have  already  been  stated,  are  obvious  from 
an  inspection  of  Figs.  74  and  78.  For  any  balanced  n-phase 
system,  the  phase  voltages  and  also  the  phase  currents  differ  in 

360 
time-phase  by degrees,  where  n  is  the  number  of  phases. 

The  complex  expressions  for  the  phase  voltages  of  an  n-phase 
system  are 

Foi  =  V  jcosO0  -  jsinO0)  (39) 

T7          T7f         360°        .    .    360°  \ 

y02  =  V  |  cos  —  -  j  sin  _  j  (40) 

0 360°  0 360°  \ 

F03  =  V  <  cos  2 j  sin  2  -     -  >  (41) 

^  H  H      } 


Tr         v\         (         ^360°        .   .     t          1N 

F0n  =  F  <  cos  (n  -  1)  -    -  -  j  sin  (n  -  1) 

L  /i  71 

where  F  is  the  magnitude  of  the  phase  voltages.     If  the  system  is 
star-connected,  the  voltage  Viz  between  line  terminals  1  and  2  is 
Fi2  =  Fi^+  Tr01 
=  -Foi  -f  F02 

360 
The  voltages  Foi  and  Fo2  are  -  -  degrees  apart  in  time-phase. 

Therefore  —Foi  and  Fo2  are  ( 180  -  —  Vdegrees  apart  in  phase. 
Hence  for  a  balanced  system,  Viz  is  equal  to  the  vector  sum  of  two 
equal  voltages  which  differ  in  phase  by  ( 180 j  degrees. 

i    /  ^firt°\ 

Fi2  =  2F  cos  £  (180°  -  -    - )  in  magnitude. 

2i  \  71     I 

1  /  ^fiO°\ 

Vune  =  2F8tor  cos  5  (180°  -  -    -)  in  magnitude.         (42) 

2  V  fi    I 

A  similar  relation  can  be  shown  to  exist  between  the  magni- 
tudes of  the  line  and  phase  currents  of  a  balanced  mesh-con- 
nected system.  For  a  mesh-connected  system 

1  /  ^fiO°\ 

I  line  =  21  mesh  cos  £(  180°  -  -    -)ia  magnitude.         (43) 
z  \  n   i 

The  following  relations  between  the  magnitudes  of  the  line 
and  phase  currents  and  between  the  line  and  phase  voltages  for 
balanced  systems  may  be  obtained  from  equations  (42)  and  (43) 
and  the  general  statements  that  precede. 


266 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


Number 
of 
phases 

Star  connection 

Mesh  connection 

Line  current 
equals 

Line  voltage 
equals 

Line  current 
equals 

Line  voltage 
equals 

3 

Iphase 

V3  V  phase 

V  3  Iphase 

'  phase 

4 

Iphase 

V2  Vphase 

V2  Iphase 

Vphase 

6 

Iphase 

'  phase 

Iphase 

Vphase 

12 

Iphase 

0.518  Vphase 

0.518  Iphase 

Vphase 

Example. — A  certain  experimental,  60-cycle,  alternating- 
current  generator  has  six  identical  armature  windings  which  are 
displaced  60  electrical  degrees  from  one  another.  When  the 
alternator  is  driven  at  rated  speed  and  has  normal  excitation,  the 
voltage  generated  in  each  winding  is  50  volts.  What  are  the  line 
and  phase  voltages  of  this  alternator  when  the  armature  windings 
are  connected  for  six-phase,  for  three-phase  and  for  single-phase? 

The  complex  expressions  for  the  voltages  in  the  armature 
windings  may  be  written 

Foi  =  50  (cos  0°  -  j  sin  0°)  =  50.0  -  JO 
F02  =  50  (cos  60°  -  j  sin  60°)  =  25.0  -  J43.3 
Fos  =  50  (cos  120°  -  j  sin  120°)  =  -25.0  -  J43.3 
F04  =  50  (cos  180°  -  j  sin  180°)  =  -50.0  -  JO 
F06  =  50  (cos  240°  -  j  sin  240°)  =  -25.0  +  j'43.3 
Foe  =  50  (cos  300°  -  j  sin  300°)  =  25.0  +  J43.3 

SIX-PHASE 

One  winding  would  be  used  for  each  phase. 

Star  Connection  Mesh  Connection 

Viz    =  Fio  4-  Fo2  Line  and  phase  voltages 

=  —  Foi  4-  Fo2  are  equal. 

=  -(50  -  JO)  4-  (25.0  -  J43.3)     Vline  =  50  volts. 
=  -25.0  -J43.3 
Fi2    =  V(-25.0p  +  (-43.3)2  =  50  volts 

Vline   =   50  VOltS 


POLYPHASE  CURRENTS  267 

THREE-PHASE 

Two  windings  would  be  used  in  series  for  each  phase.     The 
phase  voltages  would  be 

Voa  =  Foi  +  F02  =  (50.0  -  JO)  +  (25.0  -  J43.3) 

=  75.0  -  J43.3 
Vob  =  Fos  +  Fo4  =  (-25.0  -J43.3)  +  (-50.0  -JO) 

=  -75.0  -J43.3 
Voc  =  Fo5  +  Foe  =  (-25.0  +  J43.3)  +  (25.0  +  j'43.3) 

=  0+./86.6 


Star  Connection  Mesh  Connection 

b  =  Vao_  +  Vob_  Line  and  phase  vc 

=  -Voa  +  Vob  are  equal. 


=  -(75.0  -  J43.3)  VliM  =  V(75.0)2  +  (-43.3)2 

+  (-75.0  -J43.3) 

=  -150.0  -  jO  =  86.6  volts 

Vab    =  \/(-150)2  +  (0)2  =  150  volts. 

V  line  =    150  VOltS 

SINGLE-PHASE 

If  all  six  windings  were  connected  in  series,  their  resultant 
voltage  would  be  zero.  For  single-phase,  the  windings  may  be 
divided  into  two  groups  of  three  each.  The  windings  in  each 
group  would  be  connected  in  series  and  then  the  two  groups 
paralleled. 

First  group      V'  =  Voi  +  Fo2  +  Foa 

=  (50.0  -  JO)   +  (25.0   -  J43.3) 

+  (-25.0-  J43.3) 
=  50.0  -  J86.6 

Second  group  V"  =  F04  +  Fos  +  Foe 

=  (-50.0  -JO)  +  (-25.0  +  J43.3) 

+  (25.0  +J43.3) 
=  -50.0+J86.6 

The  voltages  V'  and  V"  are  equal  in  magnitude  but  opposite 
in  phase.  By  reversing  the  connections  of  one  group  of  windings, 
the  voltages  in  the  two  groups  of  windings  could  be  brought  into 


268  PRINCIPLES  OF  ALTERNATING  CURRENTS 

phase.     The  two  groups  of  windings  could  then  be  paralleled, 
giving  a  single-phase  alternator  with  a  voltage  of 


-  50.0)2  +  (86.6)  2 
=  100  volts 

and  a  current  capacity  of  twice  the  current  capacity  of  each 
winding. 

Example  of  a  Balanced  A-connected  Load.  —  Three  equal 
impedances,  each  having  a  resistance  of  10  ohms  and  an  inductive 
reactance  of  15  ohms  at  60  cycles,  are  connected  in  delta  across  a 
balanced,  three-phase,  230-volt  circuit.  What  is  the  line  current 
and  what  is  the  total  power  absorbed  by  the  load? 

The  magnitude  of  the  phase  currents  is 

j  *   phase 


" 


230  230 


V(10)2  H-   (15)2  ~  18.03 
=  12.75  amperes 

I  line    =    A/3    X    12.75 

=  22.08  amperes. 

*   phase  \-l  phase)      s\,   'phase 

=  (12.75)  2  X  10 
=  1626  watts. 

Ptotal    =   3    X  Pphase 

•=  3  X  1626 
=  4878  watts. 

Example  of  an  Unbalanced  A-connected  Load.  —  Three  im- 
pedances, z  12,  223  and  23i,  are  connected  in  delta  across  the  lines 
1-2,  2-3  and  3-1  respectively  of  a  220-  volt,  three-phase,  60-cycle 
circuit  having  balanced  voltages.  Each  of  the  impedances  has 
a  resistance  of  5  ohms.  Impedances  212  and  z23  have  inductive 
reactances  at  60  cycles  of  5  and  10  ohms  respectively.  Im- 
pedance 23i  has  a  capacitive  reactance  at  60  cycles  of  10  ohms. 
If  the  phase  order  of  the  voltages,  Fi2,  F23  and  F3i,  between  the 
terminals  1-2,  2-3  and  3-1  of  the  circuit  is  clockwise,  i.e.,  if 
7i2  leads  F23  and  723  leads  73i,  what  will  be  the  three  line  cur- 
rents? What  will  be  the  total  power  consumed  by  the  delta- 
connected  load? 


POLYPHASE  CURRENTS  269 

A  diagram  of  connections  is  shown  in  Fig.  79. 

Take  the  voltage  Viz  between  lines  1  and  2  as  the  axis  of 
reference.  The  complex  expressions  for  the  three  line  voltages 
will  then  be 

Vu  =  220  (1  +  JO)  =  220  +  JO 


F23  =  220     ~     -j  =  -110  -  J190.5 


?ai  =  220     -     +j  =  -110  +  J190.5 


-. 

FIG.  79. 

The  phase  currents,  i.e.,  the  currents  in  the  branches  of  the 
A-connected  load,  may  be  found  in  complex  by  dividing  the 
phase  voltages  in  complex  by  the  impedances  also  expressed  in 
complex.  The  phase  currents  are 

220  +  JO 


/12= 


= 


=  22  -  J22 

-110  -  J190.5 


5+jlO 
-  19.64  +jl.  18 
-110+  J190.5 

5  -  J10 
=  -19.64  -  jl.18 

From  Fig.  79,  it  is  obvious  that  the  line  currents  are 

In  =  Iiz  +  In  =  /i2  —  /si 

=  (22  -  J22)  -  (-19.64  -jl.18) 
=  41.64  -  J20.82 

In  =  V(41.64)2+  (-20.82)' 
=  46.55  amperes. 


270  PRINCIPLES  OF  ALTERNATING  CURRENTS 

/2'2    =   1  23   +  1  21   =   1  23    —  1  12 

=  (-19.64  +J1.18)  -  (22  -  J22) 
=  -41.64  +J23.18 
/2'2  =  V(-41.64)2+(23.18)2 
=  47.65  amperes. 

/3'3    =    /31    +  /32=    /31    ~   /23 

=  (-19.64  -jl.18)  -  (-19.64  +  jl.18) 
=  0  -  J2.36 


+  (-2.36)2 
=  2.36  amperes. 

The  total  power  is  equal  to  the  sum  of  the  copper  losses  in  the 
three  phases. 

Piotai  =  (/i2)2  X  r12  +  (723)2  X  r23  +  (73i)2  X  r3i 

=  !(22)2  +  (22)2)  X  5  +  {(19.64)2  +  (1.18)2)  X  5 

+  {(19.64)2  +  (1.18)2)}  X  5 
=  4840  +  1936  +  1936 
=  8712  watts. 

If  the  opposite  phase  order,  i.e.,  counter-clockwise,  had  been 
assumed,  the  total  power  would  have  been  the  same  as  for  clock- 
wise phase  order  but  the  three  line  currents  would  have  been 
different. 

Balanced  and  Unbalanced  Y-connected  Loads.—  Balanced  and 
unbalanced  Y-connected  loads  can  best  be  taken  up  after  the 
application  of  Kirchhoff's  laws  to  alternating-current  circuits  has 
been  considered.  They  will  be  considered  in  the  next  chapter. 


CHAPTER  IX 

KIRCHHOFF'S   LAWS   AND    EQUIVALENT    Y-  AND  A-CONNECTED 

CIRCUITS 

Kirchhoff's  Laws. — KirchhofFs  laws  may  be  applied  to  single- 
phase  or  to  polyphase  alternating-current  circuits  as  well  as  to 
direct-current  circuits.  When  applied  to  alternating-current 
circuits,  either  instantaneous  values  of  currents  and  voltages 
must  be  used  in  the  equations,  or  all  currents  and  voltages  must 
be  considered  in  a  vector  sense  and  must  be  referred  to  some 
conveniently  chosen  axis  of  reference.  When  instantaneous 
values  are  used,  the  resulting  equations  are  algebraic  equations. 
When  vectors  are  used,  the  resulting  equations  are  vector  equa- 
tions, and  all  currents,  voltages  and  impedances  must  be  ex- 
pressed in  their  complex  form  and  all  referred  to  the  same  axis  of 
reference.  The  solution  of  the  resulting  equations  will  give  the 
unknown  quantities  in  their  complex  form. 

Kirchhoff's  laws  applied  to  an  alternating-current  circuit  may 
be  stated  as  follows: 

For  instantaneous  values : 

(a)  The  algebraic  sum  of  the  instantaneous  values  of  all  cur- 
rents flowing  towards  any  junction  point  in  a  circuit  is  zero  at 
every  instant. 

ii  +  1*2  +  t'8  +   .    .    .   +  in  =  0 

SI  *  =  0  (1) 

(b)  The  total  rise  or  fall  of  potential  at  any  instant  in  going 
around  any  closed  circuit  is  zero. 

61  +  e2  +  £3  +   .    .    .    +  en  =  0 

s;  eHse  =  2"  e/«ii  =  0  (2) 

For  vectors: 

(a)  The  vector  sum  of  all  currents  considered  towards  any  junc- 
tion point  in  a  circuit  is  zero. 

1 1  +  72  +  h  +  .   .   .  +  7n  =  0 

rj  =  o  (3) 

271 


272  PRINCIPLES  OF  ALTERNATING  CURRENTS 

(b)  The  vector  sum  of  all  the  potential  rises  or  potential  falls 
taken  in  the  same  direction  around  any  closed  circuit  is  zero. 

Ei  +  #2  +  EB  +    •    .    .    +  Er.  =  0 

S*   Erise    =    S;     Efall    =    0  (4) 

KirchholFs  laws  hold  under  all  conditions  for  instantaneous 
values,  but  when  applied  to  vectors,  sinusoidal  currents  and 
voltages  are  presupposed.  The  current  and  voltage  of  a  circuit 
can  both  be  simple  harmonic  functions  of  time  only  when  the 
resistances,  inductances  and  capacitances  of  the  circuit  are  con- 
stant and  independent  of  current  strength.  When  the  re- 
sistances, inductances  and  capacitances  of  a  circuit  are  constant, 
KirchhofFs  laws  may  be  applied  to  it  when  the  currents  and 
voltages  are  not  simple  harmonic  functions  of  time  by  resolving 
the  currents  and  voltages  into  their  Fourier  series  and  then 
considering  the  fundamentals  and  harmonics  separately.  The 
resultant  current  or  voltage  in  any  branch  may  then  be  found 
in  the  usual  way  by  taking  the  square  root  of  the  sum  of  the 
squares  of  the  root-mean-square  values  of  the  fundamental  and 
harmonics. 

When  vectors  are  used  to  represent  equivalent  sine  waves,  the 
application  of  KirchhofFs  laws  to  a  circuit  may  and  usually  will 
give  very  inaccurate  and  generally  useless  results,  especially  when 
there  is  capacitance  as  well  as  inductance  in  the  circuit. 

Before  attempting  to  solve  any  problem  involving  KirchhofFs 
laws,  a  diagram  of  connections  should  be  made.  Each  corner 

or  junction  point  of  this  diagram 
should  be  numbered  or  lettered. 
Double  subscript  notation  should 
be  used.  (See  page  248.) 

Example    of    the   Application   of 
Kirchhoff's  Laws  to  a  Simple  Three- 
phase    Circuit. — Three    impedances 
zoa  =  10  +  JO,   zcb  =   1  +  J10  and 
FlG  80  zoc  =  0  —  jlO  are  connected  in  wye 

to  the  lines  a,  b  and  c  respectively  of 
a  three-phase,  230-volt  circuit  with  balanced  voltages,  Fig.  80. 
If  the_voltage  drop  Vdb  between  lines  a  and  b  leads  the  voltage 
drop  Vbc,  what  will  be  the  three  line  currents  and  the  voltage 


KIRCHHOFF'S  LAWS  273 

drops  between  the  lines  and  common  junction  point  of  the  im- 
pedances? If  Vab  were  taken  lagging  Vbc  instead  of  leading  it, 
a  different  solution  would  result. 

If  the  voltage  Vab  is  taken  as  the  axis  of  reference,  the  vector 
expressions  for  the  three  line  voltages  will  be 

V<*  =  230  +  JO  ^  (4) 

Vbc  =  -115  -J199.2  (5) 

Vca  =  -115  +  J199.2     u  (6) 

From  KirchhofFs  laws 

Vab  =  230  +  JO  =  7«o(10  +  JO)  +  7o&(l  + ./ 10)  (7) 

Vbc  =  -115  -  J199.2  =  7U1  +  J10)  +  7OC(0  -  J10)  (8) 

Vca  =  -115  +  J199.2  =  7CO(0  -J10)  +  7oa(10  +  JO)  (9) 

7ao   +  ho   +  Jco    =    0  (10) 

From  equation  (7) 

j         230  -  7Qfe(l  +  J10) 

10 

=  23.0-  O.llob-jToo  (11) 

From  equation  (8) 

j         -115  -J199.2  -7^(1  +J10) 

-jio 

=  -jll.5  +  19.92  -  jO.l/bo  +  Ibo  (12) 

Substituting  the  values  of  7ao  and  Ico  from  equations  (11) 
and  (12)  in  equation  (10),  remembering  that  7OC  =  —  Ico,  gives 

23.0  -  O.l7o6  -  jlob  +  ho  +  jll.5  -  19.92  +  J0.l7bo  -  ho  =  0 
7ob(0.1  +  jl.l)  =  3.08  +  jll.5  (13) 

-      _  3.08+ jll.5      0.1  -jl.l 
0.1  +J1.1    X0.1  -jl.l 
=  10.62  -  jl.834  (14) 

Substituting  the  value  of  Job  from  equation  (14)  in  equation 
(7)  gives 

-      _  230  -  (10.62  -  jl.834)(l  +  jlO) 

10 
=  20.104  -  jlO.437  (15) 

18 


274  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Substituting  the  value  of  Jbo  from  equation  (14)  in  equation 
(8)  gives 

-115  -  J199.2  -  (-10.62  +  jl.834)(l  +  J10) 
Ioc  ~  -jlO 

=  9.483  -  J8.604  (16) 
Then 

lao  =  20.10  -  jlO.44  (17) 

ho  =  -10.62  +J1.834  (18) 

Ico  =  -9.483  +  J8.604  (19) 

The  sum  of  the  three  currents  Jao,  ho  and  7CO  is  zero  as  it  should 
be. 

lao    +  ho    +   ho    =     +0+JO 

yoa  _  jo(^  2ta 

a  =  (-20.10  +  J10.44)(10  +  jQ) 
=  -201.0  +  J104.4  (20) 

Vob    =    lob  Zob 

=  (10.62  -jl.83)(l  +J10) 

=  28.92  +  J104.4  (21) 

V  oc    ==    J-  oc  Zoc 

=  (9.48  -  j8.60)(0  -  jlO) 

=  -86.0  -  J94.8  (22) 

As  a  check,  the  voltages  Voa,  V0b  and  V oc  may  be  combined  to 
give  the  voltages  between  lines. 

Vob  =  Vao  +  Vob  =  201.0  -  J104.4  +  28.92  +  J104.4 

=  229.9  +  jO 
Vbc  =  Vbo  +  Voc   =  -28.92  -  J104.4  -  86.0  -  J94.8 

=  -114.9  -  J199.2 
Vca  =  Vco  +  V.a  =  86.0  +  J94.8  -  201.0  +  J104.4 

=  -115.0  -hj'199.2 

Balanced  Y-connected  Loads  Connected  across  Three-phase 
Circuits  Having  Balanced  Voltages. — When  three  equal  imped- 
ances are  connected  in  wye  across  a  three-phase  .system  whose 
voltages  are  balanced,  the  common  junction  of  the  impedances 
is  the  true  neutral  point  of  the  three-phase  system.  The  cur- 
rents may  therefore  be  found  both  in  phase  and  in  magnitude 
by  dividing  the  wye  voltages  of  the  system  by  the  impedances, 
both  voltages  and  impedances  being  expressed  in  complex. 


KIRCHHOFF'S  LAWS 


275 


Refer  to  Fig.  81. 

Let  Zi  =  z,  z2  =  z  and  23  =  z  be  three  equal  impedances 
connected  in  wye  across  the  terminals  of  a  three-phase  system 
having  balanced  voltages.  Let  the  line  voltages  be  V 12,  V&  and 
Fsi.  The  diagram  of  connections  is  shown  in  the  left-hand  half 
of  Fig.  81.  The  vector  diagram  of  voltages  is  shown  in  the 
right-hand  half.  Foi,  Fo2  and  Fos  are  the  phase  voltages  of  a 
balanced  Y-connected  system  having  line  voltages  equal  to  Fi2, 
F23  and  F3i.  The  voltage  Fi2,  between  lines  1  and  2,  is  taken  as 
the  axis  of  reference. 


The  vector  expressions  for  the  three  line  voltages  are 
Fi2  -  F  (1  .+  JO) 


(23) 


(25) 


where  F  is  the  magnitude  of  the  voltages  between  the  lines  1-2, 
2-3  and  3-1. 

Applying  KirchhofTs  laws  to  the  circuit  gives 

(26) 


02 


Fl2  =  /102  +  / 
F23  =  /202  +  /03  2 
F31  =  /302  +  /Ol  Z 
Joi  +  /02+  /03  =  0 


(27) 
(28) 
(29) 


276  PRINCIPLES  OF  ALTERNATING  CURRENTS 

From  equations  (26)  and  (27) 

Vi*  ~  7o2  =       ~r 


•7  T^23    —    /20   g  T^23          -7  /0.,x 

103    =  -  -  -      —  =     —  —  "~    120  (61) 

Substituting  the  values  of  7i0  and  703  from  equations   (30) 
and  (31)  in  equation  (29)  gives 

^12      ,     7          ,     7  ,      F23  7 

—  —  -  --  r  ^02   +  1  02    +  -=-     -    120    = 

g 


Replacing    Viz   and    F23    by   their   vector   expressions   from 
equations  (23)  and  (24)  gives 


1 
=  —     (cos  30°  +  j  sin  30°)  X  _  (33) 


^ V(cos2  30°  -f  sin2  30° 

I  line  =  —  

V 

=  —-T=—  in  magnitude,  (34) 

where  V  is  the  magnitude  of  the  voltages  between  lines,  i.e.,  of 
the  line  voltages. 

The  delta  voltage  or  line  voltage  of  a  balanced  three-phase 
system  is  equal  in  magnitude  to  the  wye  voltage  multiplied  by  the 
square  root  of  three.  (See  equation  (22),  page  259.)  It  is  dis- 
placed from  the  wye  voltage  by  30  degrees.  Conversely,  the  wye 
voltage  is  equal  in  magnitude  to  the  delta  voltage  divided  by  the 
square  root  of  three,  and  is  displaced  from  the  delta  voltage  by 
30  degrees. 

By  referring  to  the  vector  diagram  in  Fig.  81,  it  will  be  seen 

V 
that   ~/    ^cos  30°  +  j  sin  30°)  is  the  wye  voltage  of  the  system 


KIRCHHOFF'S  LAWS  277 

for  phase  2.  The  current  in  phase  2  is  therefore  equal  in  magnitude 
to  and  in  phase  with  the  current  obtained  by  dividing  the  wye 
voltage  for  phase  2,  expressed  in  complex,  by  the  impedance  of 
the  phase,  also  expressed  in  complex. 

Expressions  similar  to  equations  (33)  and  (34)  may  be  found 
for  the  other  line  currents. 

It  follows  from  equations  (33)  and  (34)  that  the  phase  or  line 
currents  for  a  balanced  Y-connected  load,  which  is  connected  to  a 
three-phase  system  whose  voltages  are  balanced,  is  equal  in 
magnitude  to  the  line  voltage  divided  by  the  square  root  of  three 
and  by  the  phase  impedance.  It  is  equal  to  the  wye  voltage  of 
the  system  divided  by  the  phase  impedance.  The  vector  expres- 
sion for  the  current  in  any  phase,  such  as  phase  1,  is  equal  to  the 
wye  voltage  for  phase  1,  expressed  in  complex,  divided  by  the 
phase  impedance,  also  expressed  in  complex.  It  must  not  be 
forgotten  that  the  preceding  statements  are  true  only  for  a  bal- 
anced system.  If  either  the  voltages  or  the  impedances  are 
unbalanced,  the  statements  are  not  correct. 

Example  of  a  Balanced  Y-connected  Load.  —  Three  equal 
impedances,  each  having  a  resistance  of  10  ohms  and  an  inductive 
reactance  of  15  ohms,  are  connected  in  wye  to  a  230-volt  three- 
phase  system  whose  voltages  are  balanced  What  are  the  line 
currents  and  the  total  power  taken  by  the  load? 

The  wye  voltage  of  the  system  is 


V3 

Vy 


I  line    =   Iy=  


230 

\/3  132.8 


+  (15)*      18.03 
=  7.36  amperes. 

Total  power  =  P0  =  3  X  (IphaSe)2  X  rphase 
=  3  X  (7.36)2  X  10 
=   1625  watts. 

An  Example  Involving  Balanced  Y-and  A -connected  Loads  in 
Parallel  Across  a  Three-phase  System  Whose  Voltages  are 
Balanced. — Three  equal  impedances,  each  having  5  ohms 


278 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


resistance  and  5  ohms  inductive  reactance,  are  connected  in  wye 
across  a  230-volt,  three-phase  circuit  whose  voltages  are  balanced. 
Three  other  equal  impedances,  each  having  10  ohms  resistance 
and  5  ohms  capacitive  reactance,  are  connected  in  delta  across 
the  same  circuit.  What  is  the  resultant  line  current?  What  is 
the  total  power  taken  by  the  two  loads  in  parallel? 
The  complex  expressions  for  the  impedances  are 

ZY  =  5  +  J5 

gA  =  10  -  J5 

The  diagram  of  connections  and  a  vector  diagram  of  the  wye 
and  delta  voltages  are  shown  in  Fig.  82.  The  voltage  F0i 
between  neutral  and  line  1  is  taken  along  the  axis  of  imaginaries. 


FIG.  82. 


Since  the  load  is  balanced,  it  is  necessary  to  consider  the 
current  in  one  line  only.  According  to  Kirchhoff's  laws,  line 
current  In/  =  I0i  +  /2i  +  7ai. 

The  vector  expressions  for  the  voltages  Foi,  Viz  and  Vn  are 


=  0  +  J132.8  ' 

Vl2=    FlO   +    702=    -Foi+702 


.1\ 
J2/ 


230  ^-/2 
115  -  J199.2 


KIRCHHOFF'S  LAWS  279 


=  115+J199.2 

/21=    ~          =         I 

-115  +  J199.2 

10  -  jb 
-115+ J199.2 


10  -  j5         N  10  +  jb 
=  - 17.17 +J11.34 

Li  =  — 

115+  J199.2 

10  -  jb 

115  +  J199.2       10  +  jb 
10  -  j5  10  -f  j5 

=  1.232  +  J20.54 

j         Foi 

-/Ol    =    ^^ 

=  0  +  J132.8 

5  +  J5 
0  +  J132.8       5  -  j5 

5  +  jb     X  5  -  J5 
=  13.28  +  J13.28 

=  (13.28  +  J13.28)  +  (-17.17  +  jll.34) 

+  (1.232  +  J20.54) 
=  -  2.66  +  J45.15 
Jir  =  V(-2.66)2  f  (45.15)2 
=  45.23  amperes. 

Total  power  =  P0  =  37A2rA  +  3/K2r> 

=  3{(-17.17)2  +  (11.34)'}  X  10 

+  3j(13.28)2+  (13.28)2}  X5 
=  17,990  watts. 


280  PRINCIPLES  OF  ALTERNATING  CURRENTS 

This  problem  could  have  been  solved  somewhat  more  easily 
by  replacing  the  A-connected  load  by  its  equivalent  Y-connected 
load.  Equivalent  Y-  and  A-connected  loads  will  now  be  con- 
sidered. 

Equivalent  Unbalanced  Y-  and  A-connected  Three-phase 
Circuits. — If  an  equivalent  three-phase  system  is  defined  as  one 
which  takes  the  same  line  currents  at  the  same  line  voltages 
with  the  same  phase  relations  between  the  line  currents  and  line 
voltages  as  the  three-phase  system  it  replaces,  then  any  un- 
balanced delta  system  of  currents  may  be  replaced  by  just  one  equi- 
valent wye  system  of  currents ;  but  any  unbalanced  delta  system 
of  voltages  may  be  replaced  by  an  infinite  number  of  wye  systems 
of  voltages.  Fixing  either  the  voltage  or  the  constants  of  one  of 
the  equivalent  wye  branches  fixes  both  the  voltages  and  constants 
of  the  other  branches.  Conversely,  any  unbalanced  wye  system 
of  voltages  may  be  replaced  by  just  one  equivalent  delta  system  of 
voltages,  but  any  unbalanced  wye  system  of  currents  may  be  re- 
placed by  an  infinite  number  of  delta  systems  of  currents.  Fixing 
either  the  current  or  the  constants  of  one  of  the  equivalent  delta 
branches  fixes  both  the  currents  and  constants  of  the  other 
branches. 

If  the  above  definition  of  an  equivalent  three-phase  system  is 
further  qualified  by  the  condition  that  the  equivalent  system 
must  not  only  be  equivalent  as  a  whole  but  also  equivalent 
between  each  pair  of  mains,  i.e.,  if  any  main  is  opened  the  system 
must  take  the  same  current  from  the  remaining  two  mains  with 
the  same  phase  relation  between  this  cur- 
rent and  line  voltage  as  the  three-phase 
system  it  replaces  would  have  under  like 
conditions,  then  there  is  just  one  delta 
system  which  will  exactly  replace  a  wye 
system  and  conversely  there  is  just  one 
wye  system  which  will  exactly  replace  a 
^  delta  system.  That  any  A-connected 

system  may  be  replaced,  so  far  as  voltages 

alone  are  concerned,  by  an  infinite  number  of  Y-connected 
systems  will  be  evident  by  referring  to  Fig.  83.  Let  V ab,  Vbc 
and  Vca  be  the  voltages  of  the  A-connected  system  and  Voa 
Vob  and  Voc  the  voltages  of  a  Y-connected  system  which  has 


KIRCHHOFF'S  LAWS 


281 


the  same  line  voltages.  Obviously  any  three  wye  vectors 
having  their  ends  at  a,  b  and  c  will  replace  the  three  delta 
voltages.  The  common  point,  o,  from  which  the  three  wye 
vectors  are  drawn  may  be  anywhere  either  within  or  without 
the  delta. 

Relations  between  the  Constants  of  Unbalanced  Equivalent 
A-  and  Y-connected  Three-phase  Circuits. — Let  z's  with  primes 
be  the  impedances  of  the  branches  of  a  Y-connected  circuit  and 
let  z's  without  primes  be  the  impedances  of  the  equivalent  A-con- 
nected  circuit. 


Refer  to  Fig.  84.  Let  line  c  be  open.  For  the  A-connected 
system,  z\  will  then  be  in  parallel  with  z2  and  z3  in  series.  For 
the  Y-connected  system  z\  and  22'  will  be  in  series.  For  equi- 
valence between  lines  a  and  b  the  following  relation  must  hold 


2i(22  +  Z3)  «/•''.«/ 

2a6    =   ^ r7= i — r.    =   2i     +  Z2 


2i 


2.3 


With  line  a  open 


26c    = 


22(23 


+  22 


23 


With  line  b  open 


22) 


2l 


22  ' 


23' 


Subtracting  equation  (37)  from  equation  (35)  gives 


.   -  -   2322 

=  2-!  +ir+73 

Adding  equations  (36)  and  (38)  gives 

22      ==    ^ j - j ~ 


(35) 


(36) 


(37) 


(38) 


(39) 


282  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  impedances  z3'  and  z\  may  be  found  in  a  similar  manner. 

*'  -  mri, 
' 


All  impedances  must  be  expressed  in  their  complex  form. 

The  neutral  of  an  unbalanced  three-phase  A-connected  circuit 
may  be  considered  to  be  the  neutral  point  of  the  equivalent 
Y-connected  circuit  as  determined  by  the  impedances  z\  ,  zj 
and  23',  given  by  the  preceding  equations. 

Although  it  is  occasionally  convenient  in  certain  problems 
to  be  able  to  replace  an  unbalanced  A-connected  circuit  by  its 
equivalent  unbalanced  Y-connected  circuit,  little  is  saved,  as  a 
rule,  in  the  amount  of  time  and  labor  involved  in  obtaining  a 
complete  solution.  The  work  of  changing  the  given  constants  to 
the  constants  of  the  equivalent  circuit  is  usually  as  great  as  the 
labor  saved  in  solving  the  new  equivalent  unbalanced  circuit 
over  what  would  have  been  required  to  solve  the  original  circuit. 
The  conditions  are  very  different  when  the  loads  are  balanced, 
as  the.  transfer  from  a  balanced  delta  connection  to  the  equiva- 
lent balanced  wye  connection  or  vice  versa  may  be  made  quickly 
and  easily  merely  by  the  use  of  the  factor  3. 

Equivalent  Wye  and  Delta  Impedances  for  Balanced  Loads.— 
It  is  often  desirable,  when  solving  certain  types  of  problems  aris- 
ing in  engineering,  to  replace  the  impedances  of  a  balanced  A-con- 
nected load  by  impedances  connected  in  wye  which  will  take  the 
same  power  at  the  same  power-factor  from  the  three-phase 
mains.  It  may  also  occasionally  be  desirable  to  replace  a 
Y-connected  load  by  its  equivalent  A-connected  load.  In 
either  case,  in  order  to  retain  the  same  power-factor,  the  ratio 
of  the  resistance  to  the  reactance  for  the  equivalent  impedances 
must  be  the  same  as  for  the  impedances  they  replace.  If  this 
ratio  is  maintained,  it  is  merely  necessary,  when  substituting  a 
A-connected  load  for  a  Y-connected  load  or  vice  versa,  to  find 
three  impedances  which  will  take  the  same  line  current  at  the 
same  line  voltages  as  the  original  load. 

Suppose  a  balanced  A-connected  load  is  to  be  replaced  by  a 


KIRCHHOFF'S  LAWS  283 

balanced  Y-connected  load.     The  ratio  between  the  equivalent 
wye  (line)  and  delta  currents  of  any  balanced  three-phase  system  is 


Each  of  the  Y-connected  impedances  must,  therefore,  take 
the  square  root  of  three  times  as  much  current  as  each  of  the 
equivalent  A-connected  impedances,  but  the  voltages  impressed 
across  the  Y-connected  impedances  are  only  one  over  the  square 
root  of  three  times  as  great  as  the  voltage  impressed  across  the 
A-connected  impedances.  Since  the  current  varies  inversely 
as  the  impedance  of  a  circuit  and  directly  as  the  voltage,  each 
of  the  Y-connected  impedances  must  be 


V3      V3      3 

times  as  great  as  each  of  the  A-connected  impedances,  if  they  are 
to  take  the  same  line  current.  Conversely,  the  impedance  of 
each  branch  of  a  balanced  A-connected  load  must  be  three  times 
as  great  as  the  impedance  of  each  branch  of  the  balanced  Y- 
connected  load  it  replaces.  This  relation  between  the  imped- 
ances of  equivalent  Y-  and  A-connected  balanced  loads  may  be 
obtained  from  the  general  equations  (39),  (40)  and  (41),  pages 
281  and  282,  by  putting  zl  =  zz  =  z3. 

Example  of  the  Substitution  of  a  Balanced  Y-connected  Load 
for  a  Balanced  A-connected  Load. — The  problem  of  balanced 
Y-  and  A-connected  loads  in  parallel,  which  was  solved  on  page 
277,  will  be  solved  by  replacing  the  A-connected  load  by  its 
equivalent  Y-connected  load. 

The  constants  of  the  actual  loads  were 

ZY  =      5  +  J5 

ZA  =  10  -  J5 

The  impedance  of  the  Y-connected  load  which  will  replace  the 
A-connected  load  is 

_  ,       10       .5 

*  =y-^3 

The  actual  wye  impedance  and  the  equivalent  wye  impedance 
for  each  phase  may  be  treated  as  two  impedances  in  parallel 
across  a  voltage  which  is  equal  to  the  wye  voltage  of  the  system. 


284  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  resultant  admittance  of  the  two  impedances  in  parallel  is 

2/o  =  —  +  — ,  =  (QY  +  QY')  —  j(by  +  by') 

Zv  Zv 


10 
5  3 


+ 


(5,  +  (5,+  (:y+(|). 


=  {0.1  +0.24}  -  j{0.1  -  0.12} 
=  0.34  +  J0.02 

»e    ==    •*•  0    ==     '   to  neutral    X   2/0 

230 .        ,_ 
=  -/^X  V(0.34)2- 


=  132.8  X  0.3406 
=  45.23  amperes. 
Total  power  =  P0  =  3  X  (Vtoneutral)2  X  g0 


=  17,990  watts. 

That  the  substitution  of  an  equivalent  Y-connected  load  for  the 
actual  A-connected  load  simplifies  the  solution  of  this  problem 
is  evident. 

Another  Example  of  the  Substitution  of  an  Equivalent  Y- 
connected  Load  for  a  Balanced  A-connected  Load.—  A  balanced 
load,  consisting  of  three  equal  inductive  impedances,  each  having 
a  resistance  of  60  ohms  and  a  reactance  of  30  ohms,  is  connected 
in  delta  at  the  end  of  a  three-phase  transmission  line  which 
has  a  resistance  of  1  ohm  and  an  inductive  reactance  of  2 
ohms  per  conductor.  If  the  voltage  of  the  line  at  the  generating 
station  is  balanced  and  is  maintained  at  2300  volts  between 
conductors,  what  will  be  the  voltage  between  conductors  at  the 
load,  i.e.}  at  the  receiving  end  of  the  line?  What  is  the  efficiency 
of  transmission? 

This  is  really  a  case  of  a  Y-connected  load  in  series  with  a  A- 
connected  load,  the  line  impedances  being  the  Y-connected  load. 


KIRCHHOFF'S  LAWS 


285 


The  diagram  of  connections  is  shown  in  Fig.  85.  The  equivalent 
Y-connected  load  which  replaces  the  A-connected  load  is  shown 
dotted. 


FIG.  85. 


Let  FSF  and  Ffly  be  the  voltages  to  neutral,  i.e.,  the  wye  voltages, 
at  the  station  end  of  the  line  and  at  the  receiving  or  load  end 
respectively.  Then 

2300 


SY 


1328  volts. 


The  equivalent  Y-connected  impedances  which  will  replace  the 
actual  A-connected  impedances  are  each 


=  20  +  jlO 

Since  the  load  is  balanced,  the  neutral  point  at  the  load  and 
at  the  station  will  be  at  the  same  potential  and  no  current  will 
flow  between  them  even  if  they  are  connected.  Therefore  the 
voltage  to  neutral  at  the  generating  end  of  the.  line  may  be  as- 
sumed to  be  used  up  in  the  potential  drop  between  the  station  and 
the  neutral  point  of  the  equivalent  Y-connected  load.  This  drop 
will  be  equal  to  the  line  current  multiplied  by  the  resultant 
impedance  of  a  single  conductor  and  of  one  phase  of  the  equiva- 
lent Y-connected  load.  Therefore 


/fine    = 


SY 


286  PRINCIPLES  OF  ALTERNATING  CURRENTS 

where  ZL  =  1  +  J2  is  the  impedance  of  the  transmission  line  per 
conductor. 

1328 


I  line  = 


V(l  +  20)2  +  (2  +  10)2 

1328       KA  Qn 

^.    -  =  54.89  amperes. 


At  the  load  the  voltage  between  line  and  neutral  is 

VRY    =   IlineZRY  _  _____ 

=  54.89  X  V(20)2  +  (10)2 
=  1227  volts. 

At  the  load  the  voltage  between  lines  is 

F«A  =  A/3  X  VRY 
=  1.732  X  1227 
=  2125  volts. 

l™     TR* 


Efficiency  of  transmission  = 


TRY  _ 

=  0.952 


TRY 
20 


20  +  1 
=  0.952  X  100 
=  95.2  per  cent. 


CHAPTER  X 


HARMONICS  IN  POLYPHASE  CIRCUITS 

Relative  Magnitudes  of  Line  and  Phase  Currents  and  of  Line 
and  Phase  Voltages  of  Balanced  Polyphase  Circuits  when  the 
Currents  and  Voltages  are  not  Sinusoidal.  —  Let  001,  voz  and  vos  be 
the  instantaneous  induced  voltages  of  a  balanced  three-phase 
alternator,  where 


(1) 


voi  =  Vm\  sin  («0  +  Vm3  sin  3(o>0 

+  7m5  sin  5(««)  +  etc. 

^02  =  Vmi  sin  (tat  -  120°)  +  Vms  sin  3(co£  -  120°) 

+  Vm,  sin  5(<o*  -  120°)  +  etc.          (2) 

t>03  =  Vmi  sin  (at  -  240°)  +  7^  sin  3(co*  -  240°) 

+  Fm5  sin  5(co£  -  240°)  +  etc.         (3) 


The  fundamental  and  the  third  and  the  fifth  harmonics  in 
equations  (1),  (2)  and  (3)  are  plotted  in  Fig.  86. 

The  angular  displacement  between  any  harmonic  of  any  phase 
and  the  corresponding  harmonic  of  phase  one  is  given  in  the 
following  table. 


Phase 

Phase  displacement  in  electrical  degrees 

1st 

3rd 

5th 

7th 

9th 

llth 

1 

0° 

0° 

0° 

0° 

o- 

0° 

2 

120° 

3  X  120°  - 
360°  =0=0° 

5  X  120°  = 
600°  O  240° 

7  X  120°  = 
840°=  120° 

9  X  120°  = 
1080°  =0=  0° 

11  X  120°  = 
1320°  =0=240° 

3 

240°   I  3  X  240°  = 
720°  0  0° 

5  X  240°  - 
1200°  =0=120° 

7  X  240°  = 
1680°  =0=  240° 

9  X  240°  = 
2160°=G=0° 

11  X  240°  = 
2640°  =0=  120° 

A  phase  displacement  between  any  two  current  or  voltage 
vectors  of  like  frequency,  of  any  whole  number  of  wave  lengths, 
i.e.,  of  any  integral  number  of  times  360  degrees,  does  not  alter 
their  relative  phase. 

287 


288 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


It  will  be  seen  from  the  table  that  the  harmonics  of  triple 
frequency  are  in  phase,  as  are  all  harmonics  which  are  multiples 


Phase 


FIG.  86. 


of  this  frequency,  such  as  the  ninth,  fifteenth,  etc.     If  the  third 
harmonics  are  omitted,  as  well  as  all  multiples  of  the  third,  the 


HARMONICS  IN  POLYPHASE  CIRCUITS 


289 


phase  order  of  the  remaining  harmonics,  starting  with  the  first, 
or  fundamental,  alternates  from  the  order  1,  2,  3  to  the  order 
1,  3,  2.  If  the  fundamental  of  phase  one  leads  the  fundamental 
of  phase  two,  the  fifth  harmonic  of  phase  one  will  lag  the  fifth 
harmonic  of  phase  two.  The  phase  order  of  the  fifth,  eleventh, 
seventeenth,  etc.,  harmonics  is  opposite  to  that  of  the  funda- 
mental. The  phase  order  of  the  seventh,  thirteenth,  nineteenth 
etc.  harmonics  is  the  same  as  that  of  the  fundamental.  The 
vectors  for  the  fundamentals  and  harmonics  are  shown  in  Fig.  87. 


FIG.  87. 

Diagrams  of  wye  and  delta  connections  are  shown  in  Fig.  88. 


FIG.  88. 

Wye  Connection. — If  the  alternator  whose  phase  voltages  are 
given  by  equations  (1).  (2)  and  (3)  is  Y-connected,  its  terminal 
voltages  will  be  (see  Figs.  87  and  88). 

012  =  -0oi  +  002  =  \/3Fwi  sin  (co<  -J.500)  +  0 
+  \/3Fm6  sin  (5ut  +  150°) 

sin  (7wt  -  150°) 
etc.  (4) 


290  PRINCIPLES  OF  ALTERNATING  CURRENTS 


V23  =  ~v02  +  VK  =  V3Vmi  sin  (ut  -  120°  -  150°)  +  0 

sin  (5<oZ  +  120°  +  150°) 

sin  (7co*  -  120°  -  150°) 
+  etc.  (5) 

,i  sin  (<at  -  240°  -  150°)  +  0 
+  \X3FTO5  sin  (5<o*  +  240°  +  150°) 
+  \/3Fm7  sin  (7co*  -  240°  -  150°) 

+  etc.  (6) 

For  equations  (1)  to  (6)  inclusive,  time,  t,  is  zero  when  the 
fundamental  of  phase  1  is  zero. 

It  is  obvious,  from  equations  (4),  (5)  and  (6),  that  there  can 
be  no  harmonic  of  triple  frequency  or  any  harmonic,  such  as  the 
ninth  or  fifteenth,  whose  frequency  is  a  multiple  of  triple  fre- 
quency, in  the  line  voltage  of  a  balanced  three-phase  Y-connected 
alternator,  even  though  its  phase  voltage  contains  harmonics  of 
triple  frequency  or  multiples  of  this  frequency.  The  third 
harmonics  in  the  two  phases  connected  between  any  pair  of  line 
terminals  are  opposite  in  phase  when  considered  through  the 
windings  from  one  line  terminal  to  the  other  and  therefore  cancel. 
The  same  statement  is  true  regarding  the  ninth  harmonic  and 
other  harmonics  whose  frequency  is  a  multiple  of  triple  frequency. 

The  root-mean-square  or  effective  terminal  voltage  correspond- 
ing to  equations  (4),  (5)  and  (6)  is 

Fi,  =  723  =  T; 


23    —     V  31 


etc. 


The  root-rnean-square  or  effective  phase  voltage  is 
=  V     =7 

2  _i_  Y .  2  _i_  y    2  _i_  Y 72  4-  V    2  -f-  V  n2  -f-  etc 
2 

Obviously  the  ratio  of  line  to  phase  voltage  for  a  Y-connected 
alternator  cannot  be  equal  to  the  square  root  of  three  when  the 
phase  voltages  contain  harmonics  of  triple  frequency  or  other 
harmonics  which  are  multiples  of  this  frequency. 


HARMONICS  IN  POLYPHASE  CIRCUITS  291 

For  balanced  loads,  the  same  phase  relations  hold  for  phase 
currents  as  hold  for  phase  voltages.  The  fundamentals  and  all 
harmonics  that  may  exist  in  the  currents,  except  the  harmonics  of 
triple  frequency  or  multiples  of  this  frequency,  are  120  degrees 
apart  in  phase.  Their  vector  sum  will  therefore  be  zero,  since 
the  vector  sum  of  any  three  equal  vectors  which  are  120  degrees 
apart  in  phase  is  zero.  The  third  harmonics,  if  they  exist,  are 
all  in  phase.  Their  vector  sum  therefore  would  not  be  zero,  but 
three  times  the  third  harmonic  for  one  phase.  A  similar  state- 
ment is  true  of  any  harmonic  whose  frequency  is  a  multiple  of 
triple  frequency.  Since  the  vector  sum  of  all  currents  flowing 
towards  any  junction  point  in  a  circuit  must  be  zero,  it  is  obvious 
that  there  cannot  be  any  third  harmonic  current,  or  any  current 
whose  frequency  is  a  multiple  of  triple  frequency,  in  a  balanced 
three-phase  Y-connected  alternator  under  balanced  load  condi- 
tions, unless  the  neutral  point  of  the  load  and  the  neutral  point 
of  the  alternator  are  interconnected.  Without  the  neutral,  the 
only  way  the  vector  sum  of  the  harmonics  of  triple  frequency  and 
multiples  of  this  frequency  can  be  zero  at  the  neutral  point  is 
for  the  harmonics  themselves  to  be  zero. 

When  the  neutrals  are  interconnected,  the  neutral  conductor 
will  act  as  a  common  return  for  any  third-harmonic  phase  cur- 
rents that  may  exist  and  will  carry  a  third-harmonic  current 
equal  to  three  times  the  third-harmonic  current  per  phase,  a 
balanced  load  being  assumed.  It  will  also  act  as  a  common 
return  for  any  other  component  phase  currents  whose  frequency 
is  a  multiple  of  the  triple  frequency.  It  will  carry  three  times 
the  ninth  harmonic  phase  current  if  any  exists  and  three  times 
any  other  multiple  of  the  third  harmonic  current  which  is 
present  in  the  phases. 

So  long  as  the  load  remains  balanced,  the  neutral  conductor 
will  carry  no  fundamental  current  and  no  fifth,  seventh,  eleventh, 
etc.  harmonic  current.  The  three-phase  currents  in  each  of 
these  groups  are  equal  and  120  degrees  apart  for  balanced  con- 
ditions and  add  up  to  zero  at  the  neutral  connection. 

The  neutral  connection  between  the  load  and  the  alternator 
supplying  the  load  always  carries  the  vector  sum  of  the  phase 
currents  flowing  towards  the  neutral  point  of  the  circuit.  When 
the  load  is  unbalanced  the  neutral  connection  will  carry  a  current 


\ 

292  PRINCIPLES  OF  ALTERNATING  CURRENTS 

due  to  the  unbalancing.  It  may  carry  a  fundamental  current  or 
harmonic  currents  of  any  order. 

The  fact  that  wye  connection  for  balanced  loads,  without  neutral 
connection  between  the  load  and  the  source  of  power,  suppresses 
any  third  harmonic  current  which  might  otherwise  exist  is  of 
considerable  importance  under  certain  conditions.  When  a  third 
harmonic  current  is  necessary  for  the  successful  operation  of  a 
system,  a  wye  connection,  which  has  no  neutral  conductor  connect- 
ing the  neutral  of  the  load  to  the  neutral  of  the  source  of  power, 
cannot  be  used.  In  other  cases  where  it  is  desirable  to  suppress 
a  third  harmonic  current,  wye  connection  is  highly  desirable. 

Delta  Connection.  —  There  can  be  no  third  harmonic  or  multi- 
ples of  the  third  harmonic  in  the  terminal  voltage  of  a  A-con- 
nected  alternator.  In  case  any  third  harmonics  exist  in  the 
phase  voltages  of  a  A-connected  alternator,  they  will  be  in  phase 
around  the  closed  delta  formed  by  the  armature  windings  and 
are  short-circuited  therefore.  The  same  statement  is  true 
regarding  the  ninth-harmonic  current  and  any  other  multiple 
of  the  third-harmonic  current. 

If  E3  is  the  effective  value  of  the  third-harmonic  voltage  in 
each  phase  of  a  three-phase  A-connected  alternator  and  23  is  the 
phase  impedance  for  the  third-harmonic  frequency,  the  effective 
value,  73,  of  the  third-harmonic  current  short-circuited  in  the 
closed  delta  is 


Q 

32 


The  third-harmonic  voltage  in  each  phase  is  used  up  in  the 
third-harmonic  impedance  drop  in  that  phase  and  cannot  appear 
therefore  at  the  terminals  of  the  alternator.  The  possibility  of  a 
short-circuit  current  of  triple  frequency  in  the  armature  of  a 
A-connected  alternator  is  one  reason  why  delta  connection  for 
alternators  is  less  satisfactory  than  wye  connection. 

Although  there  can  be  no  harmonic  of  triple  frequency,  or  any 
multiple  of  this  frequency,  in  the  terminal  voltage  of  either  a  Y- 
or  A-connected  three-phase  alternator,  there  may  be  a  harmonic 
of  triple  frequency  and  harmonics  whose  frequencies  are  multi- 
ples of  this  frequency  between  the  line  terminals  and  neutral 
connection  of  a  Y-connected  alternator. 


HARMONICS  IN  POLYPHASE  CIRCUITS  293 

There  can  be  no  harmonic  currents  of  triple  frequency  or 
multiples  of  this  frequency  in  the  conductors  connecting  a 
balanced  A-connected  load  to  a  balanced  A-connected  alternator, 
since  those  harmonics,  evenJLilieyjejdsted  in  the  load,  would  be 
in  phase  aroun^t^^^^.a  and  their  sum  flowing  towards  any 
junction  pomt  between  a  line  and  two  phases  would  therefore 
be  zero.  Any  third  harmonics,  ninth  harmonics  or  other  multi- 
ples of  the  third  harmonic  that  existed  in  the  load  would  merely 
circulate  around  the  closed  delta  of  the  load,  without  appear- 
ing on  the  lines. 

The  line  and  the  phase  current  of  a  balanced  Y-connected 
alternator  or  load  are  obviously  the  same  both  in  magnitude  and 
in  phase.  For  a  A-connected  alternator,  the  current  in  any  line 
is  the  vector  sum  of  the  phase  currents  in  the  two  adjoining 
phases,  both  phase  currents  being  considered  towards  the  junc- 
tion point.  Referring  to  Fig.  88,  page  289,  the  current  in  line  1  is 

/I'l    =    /10_+  /03_ 

=     -/Ol    +  /03 

The  phase  relations  given  in  the  table  on  page  287  for  the 
harmonics  in  the  phase  voltages  of  a  balanced  three-phase 
alternator  also  hold  for  the  phase  relations  of  the  phase  currents 
in  any  balanced  three-phase  alternator  or  load.  In  practice 
many  of  the  harmonics  may  be  missing  or  too  small  in  magnitude 
to  be  considered.  In  certain  cases,  however,  some  of  the  har- 
monics may  be  large.  This  is  especially  true  for  inductive 
loads  containing  iron  which  is  worked  at  high  saturation.  The 
current  taken  by  such  a  circuit  will  be  non-sinusoidal  and  will 
contain  a  very  marked  third  harmonic  even  though  the  impressed 
voltage  be  sinusoidal  (see  page  231). 

If  the  phases  marked  01,  02  and  03,  Fig.  88,  page  289, 
correspond  to  phases  1,  2  and  3  respectively  in  the  table,  the  in- 
stantaneous line  currents  zYi,  iz'z  and  iyz  of  a  A-connected  alter- 
nator, which  contains  odd  harmonics  of  all  orders  in  its  phase 
voltages,  will  be 

tVi  =   -*'oi  +  i<u  =  V3Imi  sin  (at  -  8l  -  150°)  +  0 
+  A/3/ms  sin  (5o>*  -  06  +  150°) 
+  \/3/m7  sin  (lut  -  67  -  150°) 

+  etc.  (9) 


294  PRINCIPLES  OF  ALTERNATING  CURRENTS 

i2>2  =  -i<>2  +  ioi  =  V3Iml  sin  (ut  -  120°  -  61  -  150°)  +  0 
+  \/3/m5_sin  (5co£  +  120°  -  05  +  150°) 
+  \/3/w7  sin  (7co*  -  120  -  67  -  150°) 
+  etc.  (10) 

i*'i  =  -im  +  i<>2  =  VSImi  sin  (ut  -  240°  -  6l  -  150°)  +  0 
+  A/3/^sin  (5o>*  +  240°  -  05  +  150°) 
+  -v/37.,7  sin  (7co*  -  240°  -  07  -  150°) 
+  etc.,  (11) 

where  the  7TO's  with  subscripts  1,  3,  5,  etc.  represent  the  maximum 
values  of  the  fundamental  and  the  harmonics.  The  angles  0  are 
the  angles  of  lag  between  the  component  currents  and  the  corre- 
sponding component  voltages.  The  angles  6  will  be  different  for 
the  fundamental  and  the  harmonics.  For  constant  inductance, 
constant  capacitance  and  constant  resistance,  the  angle  of  lag  for 
any  harmonic,  such  as  the  nth,  is 

xc 

nx^~~ 

6n  =  tan~1- 

r 

where  XL  and  xc  are  the  inductive  and  capacitive  reactances  for 
the  fundamental  frequency. 

Equivalent  Wye  and  Delta  Voltages  of  Balanced  Three-phase 
Systems  having  Non-sinusoidal  Waves. — In  general,  there  can 
be  no  third-harmonic  voltage  between  the  lines  of  any  balanced 
three-phase  circuit.  If  there  can  be  no  third  harmonic  in  the 
voltage  between  the  lines  of  a  balanced  three-phase  system, 
obviously  there  can  be  no  third  harmonic  in  the  equivalent  wye 
voltage  of  such  a  system.  The  equivalent  wye  voltage  is  there- 
fore equal  to  the  line  voltage  divided  by  the  square  root  of  three. 
When  there  is  no  harmonic  of  triple  frequency  or  any  multiple 
of  this  frequency,  it  is  evident  from  equations  (7)  and  (8), 
page  290,  that  the  ratio  of  line  to  phase  voltage  is  equal  -to  the 
square  root  of  three  for  wye  connection.  This  relation  between 
the  line  voltage  and  the  equivalent  wye  voltage  of  a  balanced 
three-phase  circuit  is  made  use  of  in  determining  the  power-factor 
of  a  balanced  three-phase  system  which  may  contain  harmonics. 

An  Example  Illustrating  the  Relations  between  Line  and  Phase 
Voltages  of  a  Three-phase  Alternator  which  may  be  Connected 
either  in  Wye  or  in  Delta. — Oscillograph  records  show  that  the 


HARMONICS  IN  POLYPHASE  CIRCUITS  295 

phase  voltage  of  a  certain  sixty-cycle,  three-phase,  Y-connected 
alternator  contains  third,  fifth  and  seventh  harmonics,  but  no 
harmonics    of   higher    order    of   appreciable    magnitude.     The 
expression  for  the  phase  voltage  is 
e  =  1880  sin  377*  +  175  sin  (1131*  -  25°) 

+  75  sin  (1885*  -  30°)  +  30  sin  (2639*  +  40°) 

If  time  is  reckoned  from  the  instant  when  the  fundamental 
of  phase  one  is  passing  through  zero  increasing  in  a  positive 
direction,  what  are  the  expressions  for  the  voltages  of  the  three 
phases?  What  are  the  expressions  for  the  three  line  voltages? 
If  the  alternator  is  reconnected  in  delta,  what  will  be  the  three 
line  voltages?  What  are  the  root-mean-square  values  of  the 
line  and  phase  voltages  for  wye  connection?  What  is  their 
ratio?  What  is  the  ratio  of  the  line  voltages  for  wye  and  for 
delta  connection? 

The  phase  voltages  are  (see  table  on  page  287). 

e01  =  1880  sin  377*  +  175  sin  (1131*  -  25°) 
+  75  sin  (1885*  -  30°) 

+  30  sin  (2639*  +  40°) 

eQ2  =  1880  sin  (377*  -  120°)  +  175  sin  (1131*  -  25°  ±  0°) 
+  75  sin  (1885*  -  30°  +  120°) 

+  30  sin  (2639*  +  40  -  120°) 

=  1880  sin  (377*  -  120°)  +  175  sin   (1131*  -  25°  ±  0°) 
+  75  sin  (1885*  +  90°) 

+  30  sin  (2639*  -  80°) 

eoa  =  1880  sin  (377*  -  240°)  +  175  sin  (1131*  -  25°  ±  0°) 
+  75  sin  (1885*  -  30°  +  240°) 

+  30  sin  (2639*  +  40°  -  240°) 
=  1880  sin  (377*  -  240°)  +  175  sin  (1131*  -  25°) 
+  75  sin  (1885*  +  210°) 

+  30  sin  (2639*  -  200°) 

For  wye  connection  (see  Figs.  87  and  88,  page  289). 
612  =  — e0i  +  e02 

=  V3  X  1880  sin  (377*  -  150°)  +  0 
+  V3  X  75  sin  (1885*  -  30°  +  150°) 

+  \/3  X  30  sin  (2639*  -f  40°  -  150°) 
=  3256  sin  (377*  -  150°) 
+  130  sin  (1885*  +  120°) 

+  52  sin  (2639*  -  110°) 


296  PRINCIPLES  OF  ALTERNATING  CURRENTS 

023  =     —  002   +   00? 

=  V3  X  1880  sin  (377*  -  120°  -  150°)  +  0 
+  A/3  X  75  sin  (1885*  -  30°  +  120°  +  150°) 

+  V3  X  30  sin  (2639*  +  40°-  120°  -  150°) 
=  3256  sin  (377*  +  90°) 
+  130  sin  (1885*  -  120°) 
+  52  sin  (2639*  +  130°) 

031  =     —  003    +   001 

=  V3  X  1880  sin  (377*  -  240°  -  150°)  +  0 
+  \/3  X  75  sin  (1885*  -  30°  +  240°  +  150°) 

+  A/3  X  30  sin  (2639*  +  40°  -  240°  -  150°) 
=  3256  sin  (377*  -  30°) 
+  130  sin  (1885*  +  0°) 
+  52  sin  (2639*  +  10°) 

The  root-mean-square  phase  voltage  is 

T7  /(I880)2  +  (175)  2  +  (75)  2  -K30P 

y  phase  vwy  w  —  A/  2 

=  1336  volts. 
The  root-mean-square  line  voltage  for  wye  connection  is 


=  2305  volts. 

The  ratio  of  the  line  and  phase  voltages  for  wye  connection  is 
Vune  (wye)    =2305  = 
Vphasc  (wye)       1336 

This  ratio  would  be  exactly  equal  to  the  square  root  of  three 
if  there  were  no  third  harmonic  in  the  phase  voltage. 

For  delta  connection  the  third  harmonic  phase  voltage  is 
short-circuited  in  the  closed  circuit  formed  by  the  A-connected 
armature  winding,  and  therefore  cannot  appear  between  the  line 
terminals.  If  the  delta  connection  is  made  as  shown  in  Fig.  88, 
page  289,  the  line  voltages  will  be 

021  =  1880  sin  377*  +  75  sin  (1885*  -  30°) 

+  30  sin  (2639*  +  40°) 
eu  =  1880  sin  (377*  -  240°)  +  75  sin  (1885*  +  210°) 

+  30  sin  (2639*  -  200°) 

032  =  1880  sin  (377*  -  120°)  +  75  sin  (1885*  +  90°) 
+  30  sin  (2639*  -  80°) 


HARMONICS  IN  POLYPHASE  CIRCUITS  297 

The  root-mean-square   line  voltage  for  delta  connection  is 


V*.  (delta)  = 

=    1331. 

The  ratio  of  the  root-mean-square  line  voltages  for  wye  and 
delta  connections  is 

Vune  (wye  connection)  =  2305  =  =      /- 

Vnne  (delta  connection)       1331  "  V  c 

This  ratio  is  equal  to  the  square  root  of  three  as  it  must  be 
since  neither  voltage  contains  the  third  harmonic. 

An  Example  Involving  Harmonics  in  a  Balanced  Three-phase 
Circuit.  —  Three  identical  non-inductive  resistances,  when  con- 
nected in  delta  across  the  terminals  of  a  balanced  Y-connected 
alternator,  consume  12,000  watts.  When  these  same  resistances 
are  connected  in  wye  across  the  same  alternator,  they  consume 
4,750  watts  when  their  common  connection,  i.e.,  their  neutral 
point,  is  connected  to  the  neutral  of  the  alternator.  Under  this 
condition  the  current  in  the  neutral  is  15  amperes.  The  voltage 
of  the  alternator  is  assumed  to  be  the  same  in  both  cases.  What 
are  the  root-mean-square  values  of  the  voltages  between  the  line 
terminals  of  the  alternator  and  between  its  line  terminals  and 
neutral  point? 

If  there  were  no  harmonics  of  triple  frequency  or  multiples  of 
this  frequency  present  in  the  voltage  of  the  alternator,  the  power 
consumed  by  the  resistances  when  connected  in  wye  would  be 
equal  to  one  third  of  the  power  they  consume  when  connected  in 
delta  across  the  same  voltage.  This  follows  from  the  fact  that 
power  in  a  circuit  is 


When  there  is  only  pure  resistance,  i.e.,  when  the  inductance 
and  capacitance  are  each  zero,  the  expression  for  power  reduces 
to  the  voltage  squared  divided  by  the  resistance.  The  expression 

P  =  V*g 

applies  to  one  particular  harmonic,  that  for  which  the  conduc- 
tance, g,  is  computed.  The  conductance  will  have  a  different 
value  for  the  fundamental  and  each  harmonic,  except  when  the 


298  PRINCIPLES  OF  ALTERNATING  CURRENTS 

inductance  and  capacitance  are  each  zero  and  the  resistance  is 
constant.  If  the  inductance  and  capacitance  are  each  zero  and 
the  resistance  is  assumed  to  be  constant,  the  expression 


for  the  power  consumed  by  a  circuit  which  has  no  reactance 

holds  for  all  harmonics  and  fundamental,  provided  there  are  no 

harmonics  in  the  voltage  which  are  not  present  in  the  current. 

Power  in  delta  _  (Delta  voltage)2 

Power  in  wye         (Wye  voltage)2 

If  there  were  no  third  harmonic  or  multiples  of  the  third  present 
in  the  voltage,  the  ratio  of  delta  voltage  squared  to  wye  voltage 
squared  would  be  (\/3)2  =  3. 

Therefore,  if  no  harmonics  of  triple  frequency  or  multiples  of 
this  frequency  were  present  in  the  current  when  the  resistances 
were  connected  in  wye,  they  would  consume 

=  4)ooO  watts. 


The  difference  between  the  actual  power  consumed  by  the 
three  resistances  when  Y-connected,  with  their  neutral  point  and 
the  neutral  point  of  the  alternator  interconnected,  and  4,000 
watts  must  be  due  to  harmonic  currents  of  triple  frequency  or 
multiples  of  this  frequency  which  return  on  the  neutral. 

The  harmonic  currents  of  triple  frequency  and  multiples  of 
this  frequency  in  each  of  the  three  resistances  when  Y-connected 
must  be  equal  to  one-third  the  neutral  current. 

15 

-^-  =  5  amperes. 
o 

The  copper  loss,  i.e.,  the  Pr  loss,  due  to  harmonic  currents  of 
triple  frequency  and  multiples  of  this  frequency  in  each  of  the 
resistances  when  Y-connected,  is 

4,750  -  4,000      750 

—  ^—      -  =  -5-  =  250  watts. 

o  o 

Therefore 

(5)2  X  r  =  250 

r  =  10  ohms  per  resistance  unit. 


HARMONICS  IN  POLYPHASE  CIRCUITS  299 

When  the  resistances  are  A-connected  the  power  consumed, 
which  is  12,000  watts,  or  4,000  watts  per  resistance  unit,  must 
be  due  to  a  current  that  does  not  contain  any  harmonics  of  triple 
frequency  or  multiples  of  this  frequency.  This  current  must  be 

I A    (\f\f\ 

I  =  J~p  =  20  amperes. 

Therefore 

V.  =  20  X  10  =  200  volts. 


Fr  vV(v^*+(* 

=  125.8  volts. 

Harmonics  in  Balanced  Four -phase  Circuits. — There  can  be 
no  harmonic  of  triple  frequency  or  any  multiple  of  this  frequency 
in  the  line  voltage  of  a  balanced  three-phase  alternator  or  circuit. 
With  a  four-phase  alternator  or  circuit,  the  third  harmonic  and 
other  harmonics  whose  frequency  is  a  multiple  of  triple  frequency 
are  not  cut  out  from  the  line  voltages.  Odd  harmonics  of  any 
order  may  appear  in  both  the  line  and  the  phase  voltage  for 
four-phase  connection.  None  is  suppressed  by  this  connection. 
The  ratio  of  the  magnitudes  of  the  fundamentals  in  the  line 
and  phase  voltages  of  a  four-phase,  star-connected  alternator  is 
equal  to  the  square  root  of  two.  The  ratio  of  the  line  and  phase 
voltages  for  each  harmonic  that  may  be  present  is  also  equal  to 
the  square  root  of  two. 

Let  the  instantaneous  values  of  the  phase  voltages  of  a  four- 
phase  alternator  be  given  by  the  following  equations : 
VQI  =  Vmi  sin  &t  -\-  Vms  sin  3co£ 

+  Vms  sin  5co£  -f  Vm7  sin  Tut  (12) 

H-etc. 
"02  =  Vmi  sin  (co*  -  90°)  +  Vms  sin  3(«*  -  90°) 

+  Vms  sin  5(cof  -  90°)  +  Vml  sin  7(co£  -  90°) 

+  etc.  (13) 

"os  =  Vmi  sin  (ut  -  180°)+  Vm3  sin  3(o>*  -  180°) 

+  Vms  sin  5(co*  -  180°)  +  Vml  sin  7(ut  -  180°) 

+  etc.  (14) 

"04  =  Vmi  sin  (ut  -  270°)  +  Vm3  sin  3  (<at  -  270°) 

+  Vms  sin  5M  -  270°)  +  Vm7  sin  7(o>*  -  270°) 
+  etc.  (15) 


300  PRINCIPLES  OF  ALTERNATING  CURRENTS 


Phase 


FIG.  89. 


HARMONICS  IN  POLYPHASE  CIRCUITS 


301 


The  Vm's  with  the  subscripts  1,  3,  5  and  7  are  the  maximum 
values  of  the  phase  voltages. 

The  fundamentals  and  third  harmonics  in  equations  (12),  (13), 
(14)  and  (15)  are  plotted  in  Fig.  89. 

The  angular  displacement  between  any  harmonic  in  any  phase 
and  the  corresponding  harmonic  in  phase  one  is  given  in  the 
following  table. 


Phase  displacement  in  electrical  degrees 


1st 

3rd 

5th 

7th 

9th 

llth 

1 

0° 

0° 

0° 

0° 

0° 

0° 

2 

90° 

3  X  90°  = 
270° 

5  X  90°  = 
450°  0  90° 

7  X  90°  = 
630°  0  270° 

9  X  90°  = 
810C090° 

11  X  90°  = 
990°  0  270° 

3 

180° 

3  X  180°  = 
540°  01  80° 

5  X  180°  = 
900°  0180° 

7  X  180°  = 
1260°  0180° 

9  X  180°  = 
1620°  0180° 

11  X  180°  = 
1980°  0  180° 

4 

270° 

3  X  270°  = 
8  10°  090° 

5  X  270°  = 
1350°  0270° 

7  X  270°  = 
1890°  090° 

9  X  270°  = 
2430°  0  270° 

11  X  270°  = 
2970°  0  90° 

It  will  be  seen  from  the  preceding  table  that  the  harmonics  of 
like  frequency  in  the  four  phases  of  a  balanced  four-phase  alter- 
nator or  balanced  four-phase  load  are  ninety  degrees  apart  in 
time-phase.  It  should  also  be  noticed  that  the  phase  order  of  the 


fundamentals  and  harmonics  of  different  frequencies  alternates 
from  the  phase  order  1-2-3-4  to  the  phase  order  1-4-3-2.  One 
phase  order  is  opposite  to  the  other.  See  Fig.  90. 

The  vectors  F0i,  F02,  F03  and  F04,  for  the  fundamentals  and 


302 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


harmonics  for  a  balanced  four-phase  alternator  or  load,  are  shown 
in  Fig.  90. 

Star  and  mesh  four-phase  connections  are  shown  diagrammati- 
cally  in  Fig.  91. 


FIG.  91. 

The  line  voltages  for  the  fundamental  or  any  harmonic,  for 
star  connection,  are 

Fi2  =    -  Foi  +  F02  (16) 

F23  =   -7o2  +  Fo3  (17) 

F34    =     -F03   +   704  (18) 

F4i  =  -Fo4  +  Foi  (19) 

Since  the  component  voltages  of  like  frequency  in  the  four 
phases  differ  in  time-phase  by  90  degrees,  the  line  voltage  for 
the  fundamental  or  the  line  voltage  for  any  harmonic  is 

Vure    =    2Vphase   COS  45° 

=  \/2  Vphase  (20) 

where  Vune  is  the  magnitude  of  the  voltage  between  any  pair  of 
adjacent  lines  for  the  fundamental  or  any  harmonic  and  Vphase  is 
the  magnitude  of  the  corresponding  voltage  in  the  phases. 

The  instantaneous  voltages  between  lines,  i.e.,  line  voltages, 
for  star  connection,  corresponding  to  the  phase  voltages  given  in 
equations  (12),  (13),  (14)  and  (15)  are  (see  Figs.  90  and  91). 

Vi2  =  -VQI  +  v02  =  \/2Vml  sin  (otf  -  135°) 

sin  (3«*  +  135°) 

sin  (5coZ  -  135°) 
\/2Vm7  sin  (7at  +  135°) 
+  etc.  (21) 


HARMONICS  IN  POLYPHASE  CIRCUITS  303 

sin  (at  -  90°  -  135°) 


sin  (3co*  +  90°  +  135°) 
sin  (5co*  -  90°  -  135°) 

sin  (7wt  +  90°  +  135°) 
+  etc.  (22) 

V34  =  _V03  +  Vo4  =  ^/2Vml  sin  (at  -  180°  -  135°) 

sin  (3a>t  +  180°  +  135°) 

sin  (5co<  -  180°  -  135°) 
+  A/2  Fm7  sin  (7co^  +  180°  +  135°) 
+  etc.  (23) 

sin  (w'  -  270°  -  135°) 

sin  (3^  +  270°  +  135°) 

sin  (5co*  -  270°  -  135°) 
+  V~2Vm7  sin  (7co^  +  270°  +  135°) 
+  etc.  (24) 

The  root-mean-square  values  of  the  line  and  phase  voltages 
for  star  connection  are 


etc. 


17     ,2  _|_  V       2     I      V       2117       21     Af« 
'   ml       I      r  m3  '  w»5     HT    '   w»7     JT  "to. 

~T~ 
^  =  V2  (27) 

'   phase 

The  ratio  of  line  to  phase  root-mean-square  voltage  for  a  four- 
phase,  balanced,  star-connected  alternator  or  circuit  is  always 
equal  to  the  square  root  of  two.  This  is  entirely  independent  of 
wave  form.  Although  the  line  and  phase  voltages  will  contain 
like  harmonics  in  the  same  relative  magnitudes,  the  wave  forms 
of  the  two  voltages  will  not  be  alike.  The  relative  phase  dis- 
placements of  the  harmonics  in  the  two  voltages  will  be  different, 
as  will  be  seen  by  comparing  equations  (21),  (22),  (23)  and  (24) 
with  equations  (12),  (13),  (14)  and  (15)  on  page  299. 

The  line  and  phase  voltages  of  a  mesh-connected,  four-phase 


304  PRINCIPLES  OF  ALTERNATING  CURRENTS 

alternator  or  circuit  are  the  same  both  in  magnitude  and  wave 
form,  since  no  harmonic  is  short-circuited  in  a  mesh-connected 
four-phase  circuit  and  thus  eliminated  from  the  terminal  voltage. 

For  balanced  conditions,  the  same  relations  hold  between  line 
and  phase  currents  for  four-phase  mesh  connection  as  hold  between 
line  and  phase  voltages  for  a  balanced,  four-phase,  star-connected 
circuit. 

If  the  neutral  of  a  balanced,  four-phase,  star-connected  load  is 
connected  to  the  neutral  of  the  source  of  power  supplying  the 
load,  no  current  will  flow  in  the  neutral,  since  the  fundamentals 
of  the  phase  currents  and  all  harmonics  of  any  order  that  may 
exist  in  the  phase  currents  are  ninety  degrees  apart  in  time-phase 
and  therefore  add  up  to  zero  at  the  neutral  point.  The  vector 
sum  of  any  four  equal  vectors  which  differ  in  phase  by  ninety 
degrees  is  equal  to  zero.  The  only  current  carried  by  the  neutral 
connection  of  a  four-phase,  star-connected  system  is  that  due  to 
an  unbalanced  load. 

Since  the  harmonics  as  well  as  the  fundamentals  in  the  phase 
voltages  of  a  balanced  four-phase  system  are  ninety  degrees 
apart  in  time-phase,  there  can  be  no  circulatory  current  in  the 
armature  of  a  mesh-connected  four-phase  alternator,  whose 
voltages  are  balanced,  since  the  vector  sum  of  the  component 
voltages  of  any  given  frequency  will  be  zero.  This  is  different 
from  the  conditions  which  may  exist  in  the  armature  of  a  three- 
phase,  mesh  connected,  i.e.,  delta-connected  alternator.  In  this 
case  the  third  harmonics  and  all  harmonics  whose  frequency  is  a 
multiple  of  triple  frequency  are  short  circuited  in  the  closed  delta 
formed  by  the  armature  windings. 

Harmonics  in  Balanced  Six-phase  Circuits. — Although  no 
six-phase  alternators  are  built,  there  is  a  demand  for  a  consider- 
able amount  of  six-phase  power  for  the  operation  of  rotary  con- 
verters, or  synchronous  converters,  as  they  are  sometimes  called. 
This  is  always  obtained  from  three-phase  systems  by  means  of 
ordinary  static  transformers  connected  for  three-phase  to  six- 
phase  transformation.  In  most  of  the  larger  cities,  alternating- 
current  generation  and  transmission  of  power  are  used,  but 
direct-current  distribution  is  usually  employed  in  the  business  and 
thickly  settled  districts.  The  rotary  converter  is  also  used 
in  connection  with  most  street  railways  operating  on  direct 


HARMONICS  IN  POLYPHASE  CIRCUITS  305 

current.  The  rotary  converter  is  the  chief  connecting 
link  between  the  alternating-current  and  the  direct-current 
systems.  The  rotary  converter  is  essentially  a  direct- 
current  generator  which  has  taps  brought  out  to  slip-rings 
from  equidistant  points  on  its  armature.  It  receives  polyphase 
alternating-current  power  through  the  slip-rings  and  delivers 
direct-current  power  from  the  direct-current  commutator.  On 
the  alternating-current  side  it  acts  as  a  synchronous  motor  and 
on  the  direct-current  side  it  acts  as  a  direct-current  generator. 
A  single  armature  with  a  single  armature  winding  serves  for  both 
motor  and  generator  action.  The  armature  winding  may  be 
considered  to  carry  a  current  which  is  the  difference  of  the 
alternating-current  supplied  and  the  direct-current  delivered. 
On  account  of  the  armature  current  being  equal  to  the  differ- 
ence between  the  two  currents,  a  greater  direct-current  out- 
put may  be  obtained  for  a  fixed  average  armature  copper  loss 
than  could  be  obtained  were  the  machine  driven  mechanically 
and  loaded  as  a  direct-current  generator.  For  a  fixed  amount  of 
power  converted,  the  efficiency  of  a  rotary  converter  is  higher  and 
the  cost  is  less  than  for  a  motor-generator  which  would  do  the 
same  work.  The  gain  in  output  and  efficiency  increases  with 
the  number  of  phases  used.  For  three-phase  operation  at  unity 
power-factor,  the  output  of  a  rotary  converter  is  about  thirty  per 
cent  greater  than  could  be  obtained  from  the  same  machine 
operated  as  a  direct-current  generator.  For  six-phase  operation 
it  is  nearly  double.  There  is  not  a  proportionate  gain  in  going 
to  twelve-phase  operation,  although  it  may  pay  to  build  very  large 
rotary  converters  for  this  number  of  phases. 

On  account  of  the  importance  of  the  rotary  converter  as  a 
connecting  link  between  alternating-current  transmission  and 
direct-current  distribution,  it  is  worth  while  to  give  some  con- 
sideration to  the  relations  among  the  harmonics  in  six-phase 
circuits. 

Six-phase  systems  are  always  derived  from  polyphase  systems, 
usually  three-phase,  by  means  of  transformers.  A  six-phase 
system  may  be  considered  to  be  two  three-phase  systems  which 
are  superposed,  with  the  voltages  of  one  reversed  with  respect 
to  those  of  the  other.  This  reversal  may  be  obtained  by  merely 
reversing  the  connections  of  the  secondary  windings  of  the  trans- 
20 


306 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


formers  supplying  one  of  the  superposed  systems.  Consider 
Fig.  92.  This  figure  shows  the  vectors  representing  the  funda- 
mental voltages  of  a  six-phase  system. 


FIG.  92. 

The  two  superposed  three-phase  systems,  to  which  the  six- 
phase  system  is  equivalent,  are  distinguished  in  Fig.  92  by  using 
full  lines  for  one  and  dotted  lines  for  the  other. 

Figure  93  shows  the  six  voltage  vectors  for  the  fundamentals 
of  a  six-phase  system. 


Star  Connection 


Mesh  Connection 


FIG.  94 


Either  star  or  mesh  connection  may  be  used  for  six-phase 
circuits,  but  for  the  rotary  converter,  mesh  connection  must  be 
used,  since  the  armature  winding  must  serve  to  generate  direct 
current  and  must  therefore  form  a  closed  circuit.  All  direct- 
current  armatures  must  have  closed-circuit  windings.  Star 
and  mesh  connection  are  illustrated  diagrammatically  in  Fig.  94. 

The  following  table  gives  the  phase  relations  among  the 
fundamentals  and  harmonics  in  a  balanced  six-phase  system. 


HARMONICS  IN  POLYPHASE  CIRCUITS 


307 


The  table  gives  the  angular  displacement  between  any  harmonic 
in  any  phase  with  respect  to  the  corresponding  harmonic  in  phase 
one. 


Phase 

Phase  displacement  in  electrical  degrees 

1st 

3rd 

oth 

7th 

9th 

llth 

1      0° 

0° 

0° 

0° 

0° 

0° 

2 

60° 

3  X  60°  - 
180° 

5  X  60°  = 
300° 

7  X  60°  = 
420°  =0=  60° 

9  X  60°  = 
540°  =0=  180° 

11  X  60°  = 
660°  =0=  300° 

3 

120° 

3  X  120°  = 
360°  =C=0<> 

5  X  120°  = 
600°  =0=  240° 

7  X  120°  = 
840°  0  120° 

9  X  120°  = 
1080°  =0=  0° 

11  X  120°  = 
1320°  =0=  240° 

4 

180° 

3  X  180°  = 
540°  =0=  180° 

5  X  180°  - 
900°  01  80° 

7  X  180°  = 
1260°  =0=  180° 

9  X  180°  = 
1620°  =0=1  80° 

11  X  180°  = 
1980°  0  180° 

5 

240° 

3  X  240°  - 
720°  00° 

5  X  240°  = 
1200°  =0=120° 

7  X  240°  = 
1680°  =0=240° 

9  X  240°  = 
2160°  =0=  0° 

11  X  240°  = 
2640°  =0=  120° 

6 

300° 

3  X  300°  - 
900°  =C=  180° 

5  X  300°  - 
1500°  =0=  60° 

7  X  300°  = 
2100°  =0=300° 

9  X  300°  = 
2700°  =0=  180° 

11  X  300°  = 
3300°  =0=  60° 

The  vectors  for  the  fundamental  and  for  the  harmonics  of  a 
balanced  six-phase  system  are  shown  in  Fig.  95. 

From  inspection  of  the  table  it  is  obvious  that  for  a  balanced 
six-phase  system,  all  harmonics  which  are  present,  except  those  of 


For  1st,  7th,  13th  etc. 


020406 
For  3rd,  9th,  15th  etc. 

FIG.  95. 


For  5th,  llth,  17th  etc. 


triple  frequency  and  multiples  of  this  frequency,  differ  by  sixty 
degrees  in  time-phase  among  themselves. 

The  harmonics  of  triple  frequency  and  multiples  of  this  fre- 
quency are  either  in  phase  or  in  phase  opposition  among  them- 


308  PRINCIPLES  OF  ALTERNATING  CURRENTS 

selves,  those  in  adjacent  phases  always  being  in  phase  opposition. 
Omitting  the  harmonics  of  triple  frequency  and  multiples  of 
this  frequency,  the  phase  order  of  the  others,  beginning  with  the 
first  harmonic,  i.e.,  the  fundamental,  alternates  from  the  direct 
order,  i.e.,  1-2-3-4-5-6,  to  the  reverse  order,  i.e.,  1-6-5-4-3-2. 
These  phase  relations  among  the  fundamentals  and  harmonics 
of  the  different  orders  are  what  would  be  expected  and  what 
would  necessarily  follow  from  the  phase  relations  existing 
among  the  fundamentals  and  harmonics  of  a  balanced  three- 
phase  system. 

For  a  balanced,  star-connected,  six-phase  system,  the  line 
voltage  for  the  fundamental  and  for  any  harmonic,  except  those 
of  triple  frequency  and  multiples  of  this  frequency,  is  equal  to 
the  phase  voltage  in  magnitude.  Refer  to  Figs.  94  and  95. 

7i2  =  -Foi  +  Fo2 

—  Voi  =  Vio  (=  ^04)  and  F02  are  two  equal  voltages  which  are 
120  degrees  apart  in  phase.  Therefore 

Vi2    =    Vline    =   2Vphase  COS  60° 

=  2  X  ~Vphase  =  Vphase  (28) 

For  the  harmonics  of  triple  frequency  or  multiples  of  this 
frequency  in  a  balanced  three-phase  system,  —  F0i  and  Vo2  are  in 
phase.  Therefore  for  a  balanced,  star-connected,  six-phase 
system 

Fi2  =  -Foi  +  Fo2 

Vline    =   2Vphas6  (29) 

There  can  be  no  voltages  of  triple  frequency  or  multiples  of 
this  frequency  in  the  voltages  between  alternate  line  terminals 
of  a  balanced,  star-connected,  six-phase  system.  Consider 
terminals  1  and  3.  The  voltage  between  these  terminals  for  star 
connection  is  (Fig.  94,  page  306.) 

Fl3    =     -Foi    +    F03 

By  referring  to  the  table  on  page  307  it  will  be  seen  that  the 
voltage  Fis  is  zero  for  the  harmonics  of  triple  frequency  and 
multiples  of  this  frequency. 

For  mesh  connection,  the  voltage  between  terminals  1  and  3 
is,  (Fig.  94),  page  306,_ 

Vis  =  -(Foi  +  Fo2)  (30) 


HARMONICS  IN  POLYPHASE  CIRCUITS  309 

By  referring  to  the  table,  it  will  be  seen  that  this  is  zero  for 
harmonics  of  triple  frequency  or  multiples  of  this  frequency. 

Although  there  can  be  no  harmonics  of  triple  frequency  or 
multiples  of  this  frequency  in  the  voltage  between  alternate 
terminals,  i.e.,  between  the  three-phase  terminals,  of  a  balanced 
six-phase  system,  there  may  be  harmonics  of  triple  frequency  or 
multiples  of  this  frequency  in  the  voltage  between  any  diametrical 
terminals,  such  as  1  and  4,  of  a  balanced  six-phase  system. 

Similar  relations  exist  among  the  line  and  phase  currents  of  a 
balanced,  mesh-connected,  six-phase  system  to  those  holding 
between  the  line  and  phase  voltages  of  a  balanced,  star-connected, 
six-phase  system. 

When  mesh  connection  is  used  for  an  armature,  there  can  be  no 
short-circuit  current  in  the  armature,  as  there  may  be  in  the  case 
of  three-phase  delta  connection,  since  the  vector  sum  of  all  the 
component  voltages  acting  around  a  mesh-connected,  balanced, 
six-phase  circuit  is  zero.  By  referring  to  the  table  on  page  307 
it  will  be  seen  that  the  vector  sum  of  the  six-phase  voltages 
acting  around  a  mesh-connected,  balanced,  six-phase  circuit  is 
zero  for  the  fundamental  and  for  each  harmonic. 

The  diametrical  voltage  of  a  six-phase,  mesh-connected  circuit 
between  any  two  diametrical  terminals,  such  as  terminals  1  and 
4,  is 

Fl4    =    Foi    +    F02    +    F03  (31) 

By  referring  to  the  table  on  page  307  it  will  be  seen  that  for 
harmonics  of  triple  frequency  or  multiples  of  this  frequency,  this 
is  equal  to  Fos  or  Voi. 

For  star  connection,  the  diametrical  voltage  between  any  dia- 
metrical terminals,  such  as  1  and  4,  for  balanced  conditions  is 

FH  =  -Foi  +  7o4  (32) 

By  referring  to  the  table  on  page  307  it  will  be  seen  that,  for 
the  third  harmonic  or  any  multiple  of  the  third  harmonic,  Vu  = 


Therefore,  for  mesh  connection,  the  third-harmonic,  six-phase. 
line  voltage  is  equal  to  the  third-harmonic  phase  voltage.  For 
star  connection  it  is  double  the  phase  voltage  The  same  state- 
ments hold  for  any  multiple  of  the  third.  Balanced  conditions 
are  assumed. 


310  PRINCIPLES  OF  ALTERNATING  CURRENTS 

In  general  for  a  balanced  six-phase  system,  no  harmonics  are 
suppressed  in  the  voltage  between  adjacent  terminals.  Any 
harmonic  voltage  that  exists  in  the  phase  voltage  will  be  present 
in  the  voltage  between  adjacent  terminals  The  harmonics  of 
triple  frequency  and  multiples  of  this  frequency  that  exist  in  the 
phase  voltage  are  suppressed  between  alternate  terminals,  i.e., 
between  the  three-phase  terminals.  For  mesh  connection,  the 
phase  and  line  voltages  are  identical.  For  star  connection, 
phase  and  line  voltages  are  the  same  only  when  the  phase  voltages 
contain  no  harmonics  of  triple  frequency  or  multiples  of  this 
frequency.  For  star  connection,  the  fundamental  line  and  phase 
voltages  are  equal.  The  line  and  phase  voltages  for  any  given 
harmonic,  except  the  third  or  its  multiples,  are  also  equal.  For 
the  third-harmonic  or  any  harmonic  whose  frequency  is  a  multiple 
of  triple  frequency,  the  line  voltage  for  star  connection  is 
double  the  phase  voltage.  It  is  equal  to  the  phase  voltage  for 
mesh  connection. 

An  Example  Illustrating  the  Relation  Among  the  Voltages 
of  a  Six -phase  Mesh-connected  System  Having  a  Badly  Dis- 
torted Wave  Form. — One  type  of  rotary  converter  is  purposely 
built  in  such  a  way  that  its  phase  voltage  may  be  badly  distorted 
by  changing  the  flux  distribution  in  the  air  gap  in  order  to  alter 
the  root-mean-square  alternating-current  voltage  obtained  with 
a  given  pole  flux.  In  this  way  it  is  possible  to  change  the  ratio 
of  the  alternating-  and  direct-current  voltages.  Under  certain 
conditions,  the  phase  voltage,  i.e.,  the  voltage  between  adjacent 
armature  taps,  was  found  to  be 

e  =  325  sin  377*  +  110  sin  (1131*  +  90°) 
+  50  sin  (1885J  +  50°) 

If  time  is  reckoned  from  the  instant  when  the  fundamental  in 
the  voltage  of  the  phase  between  armature  taps  1  and  2  is  zero 
and  increasing  in  a  positive  direction,  what  are : 

(a)  The  expressions  for  the  six  instantaneous  six-phase  voltages? 

(6)  The  expression  for  the  instantaneous  three-phase  voltage 
between  armature  taps  1  and  3? 

(c)  The  expression  for  the  instantaneous  diametrical  voltage 
between  armature  taps  1  and  4? 

(d)  What  are  the  root-mean-square  values  of  the  six-phase 


HARMONICS  IN  POLYPHASE  CIRCUITS  311 

voltages,  i.e.,  the  voltages  between  adjacent  armature  taps;  the 
three-phase  voltages,  i.e.,  the  voltages  between  alternate  arma- 
ture taps;  and  the  diametrical  voltages,  i.e.,  the  voltages  between 
any  two  diametrical  armature  taps  such  as  1  and  4? 

The  voltage  between  armature  taps  1  and  2  will  be  assumed  to 
lead  the  voltage  between  taps  2  and  3. 

The  armature  of  a  rotary  converter  is  mesh-connected. 

The  expressions  for  the  six  instantaneous  six-phase  voltages 
are  (see  Figs.  94  and  95,  pages  306  and  307  respectively  and  also 
the  table  on  page  307) 

en  =  elo  =  325  sin  377*  +  110  sin  (1131*  +  90°) 

+  50  sin  (1885*  +  50°) 
e23  =  e20  =  325  sin  (377*  -  60°)  +  110  sin  (1131*  -  180°  +  90°) 

+  50  sin  (1885*  -  300°  +  50°) 
e34  =  630  =  325  sin  (377*  -  120°)   +110  sin  (1131*  -  0°  +  90°) 

+  50  sin -(1885*  -  240°  +  50°) 
e45  =  640  =  325  sin  (377*  -  180°)  +  110  sin  (1131*  -  180°  +90°) 

+  50  sin  (1885*  -  180°  +  50°) 
6o6  =  6b0  =  325  sin  (377*  -  240°)  +  110  sin  (1131*  -  0°  +  90°) 

+  50  sin  (1885*  -  120°  +  50°) 
eei  =  eeo  =  325  sin  (377*  -  300°)  +  110  sin  (1131*  -  180°  +  90°) 

+  50  sin  (1885*  -  60°  +  50°) 

The  root-mean-square  value  of  the  six-phase  voltage  is 


*  six-phase    —    A/ 


(325)2  +  (110)2  +  (50)2 


=  245.2  volts. 

The  expression  for  the  instantaneous  voltage  between  armature 
taps  1  and  3  is  (see  Fig.  94,  page  306.) 

=  A/3  X  325  sin  (377*  -  30°)  +  0° 

+  V3  X  50  sin  (1885*  +  50°  +  30°) 
=  562.9  sin  (377*  -  30°)  +  86.6  sin  (1885*  +  80°) 

The  root-mean-square  value  of  the  voltage  between  armature 
taps  1  and  3  is 

v          v  /(562.9)2  +  (86.6)' 

r  13    —     ¥  three-phase    — 


2 
402.7  volts 


312  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  ratio  of  the  three-phase  and  six-phase  voltages  is 

Vthfee-phase    _  402.7      _    .,    _  .~ 

Vsix.phase        245.2  " 

When  there  are  no  third  harmonics,  or  multiples  of  the  third, 
present  in  the  six-phase  voltage,  the  ratio  of  the  three-phase  and 
six-phase  voltages  is  equal  to  1.732,  i.e.,  the  square  root  of  three. 

The  voltage  between  any  two  diametrical  armature  taps  is 
equal  to  the  vector  sum  of  three  equal  vectors  which  are  displaced 
by  equal  angles.  The  easiest  way  to  find  the  resultant  of  these 
three  vectors  is  to  add  the  first  and  third  vectors  and  then  add 
their  resultant  to  the  second  vector.  For  the  fundamental  and  all 
harmonics,  except  those  of  triple  frequency  or  a  multiple  of  this 
frequency,  the  first  and  third  vectors  are  120  degrees  apart  in 
time-phase.  The  vector  sum  of  these  is  equal  both  in  phase  and 
in  magnitude  to  the  second  vector.  The  vector  sum  of  the  three 
vectors  therefore  is  in  phase  with  the  second  vector  but  has  twice 
its  magnitude.  That  this  statement  is  true  will  be  seen  by  refer- 
ring to  Fig.  95,  page  307.  For  the  harmonic  of  triple  frequency  or 
for  harmonics  of  any  multiple  of  triple  frequency,  the  first  and 
second  vectors  are  opposite  in  phase.  (See  Fig.  95,  page  307.) 
Their  vector  sum  is  zero.  The  resultant  of  the  three  vectors  for 
the  harmonic  of  triple  frequency  or  for  harmonics  of  any  multiple 
of  triple  frequency  is  equal  in  direction  and  magnitude  to  the 
third  vector. 

It  follows  from  the  preceding  statements  that  the  expression 
for  the  instantaneous  voltage  between  diametrical  armature  taps 
1  and  4  is 

014    =    010   +  020   +  030 

=  2  X  325  sin  (377*  -  60°) 

+  110  sin  (1131*  +  90°  -  180°) 

+  2  X  50  sin  (1885*  +  50°  +  60°) 
=  650  sin  (377*  -  60°)  +  110  sin  (1131*  -  90°) 

+  100  sin  (1885*  +  110°) 

The  root-mean-square  value  of  the  voltage  between  diametrical 
armature  taps  1  and  4  is 

v      _  y  /(650)2  +  (IIP)2  +  (100)2 

*  14    ~      V  diametrical  A/  ~ 

^  471.5  volts, 


HARMONICS  IN  POLYPHASE  CIRCUITS  313 

The  ratio  of  the  diametrical  and  six-phase  voltages  is 

Vdiametrical      _  471.5    _ 

Vtit.p^        245.2  = 

When  there  are  no  harmonics  of  triple  frequency  or  multiples 
of  this  frequency  present  in  the  voltages,  the  ratio  of  the  diam- 
etrical and  six-phase  voltages  is  equal  to  2. 

It  should  be  noted  that  the  wave  forms  of  the  six-phase,  three- 
phase  and  diametrical  voltages  are  all  different. 


CHAPTER  XI 

POWER  AND  POWER-FACTOR  OF  POLYPHASE  CIRCUITS,  RELATIVE 

AMOUNTS  OF  COPPER  REQUIRED  FOR  POLYPHASE  CIRCUITS, 

POWER  MEASUREMENTS  IN  POLYPHASE  CIRCUITS 

Power  and  Power -factor  of  Balanced  Polyphase  Circuits.— 

Since  a  polyphase  alternator  has  as  many  independent  windings 
on  its  armature  as  it  has  phases,  it  is  evident  that  the  total  output 
of  such  an  alternator  must  be  equal  to  the  sum  of  the  outputs  of 
all  its  phases,  no  matter  how  they  may  be  interconnected.  In 
general,  the  power  in  any  polyphase  circuit  whatsoever  is  equal 
to  the  sum  of  the  powers  developed  in  each  phase. 

Total  power  =  P0  =  PI  +  P2  +  PS  +  etc.  (1) 

=  ZP 

where  the  P's  with  subscripts  1,  2,  3,  etc.  are  the  powers  developed 
in  phases  1,  2,  3,  etc.  respectively. 
For  a  three-phase  circuit 

Po  =  Pi  +  P2  +  P3 

=  TVi  X  (p./.)i  +  VJ*  X  (p./.)2 

+  TVs  X  (p./.)s  (2) 

Vi,  Vz  and  V3  are  the  phase  voltages.  7i,  72  and  J3  are  the 
corresponding  phase  currents,  and  (p./.)i,  (p./.) 2  and  (p./.)3 
are  the  corresponding  phase  power-factors.  For  sinusoidal  waves, 
the  power-factor  is  equal  to  the  cosine  of  the  angle  between 
the  phase  current  and  the  phase  voltage. 

The  total  volt-amperes  of  a  circuit  is  equal  to  the  sum  of  the 
volt-amperes  in  its  phases. 

Total    volt-amperes  =  Vlll  +  V2I2  +  F3/3  +  etc.  (3) 

=  SF7 

In  a  balanced  polyphase  circuit,  all  the  phase  currents  are 

360 
equal  in  magnitude  and  differ  in  phase  by  -  -  degrees,  where  n  is 

314 


POWER  AND  POWER-FACTOR  315 

the  number  of  phases.     All  phase  voltages  are  also  equal  in  mag- 

Oflfl 

nitude  and  differ  in  phase  by  —  degrees.     It  follows  for  a  bal- 

anced polyphase  circuit  that  the  phase  power-factors  must  all 
be  equal. 

For  a  balanced  circuit 

Total  power  =  P0  =  nPp  =  nVJp  X  (p.f.)P  (4) 

where  n  is  the  number  of  phases  and  Vp,  Ip  and  (p.f.)P  are  the 
phase  voltage,  the  phase  current  and  the  phase  power-factor 
respectively. 

For  a  balanced  three-phase  circuit, 

P0  =  3PP  =  3TVp  X  (p./.)*  (5) 

Total  volt-amperes  =  3VPIP  (6) 

For  a  A-connected  circuit,  the  line  and  phase  voltages  are  the 
same.  For  a  Y-connected  circuit,  the  line  and  phase  currents 
are  the  same.  For  a  balanced  circuit  having  sinusoidal  waves  of 
current  and  voltage,  the  line  current  is  equal  to  the  phase  current 
multiplied  by  the  square  root  of  three  for  delta  connection.  The 
line  voltage  is  equal  to  the  phase  voltage  multiplied  by  the  square 
root  of  three  for  wye  connection. 

Indicating  line  current  and  line  voltage  by  IL  and  VL  respec- 
tively, equations  (5)  and  (6)  for  sinusoidal  waves  and  a  balanced 
A-connected  circuit  become 

Total  power  =  P0=  3VPL  (p./.), 


=  V3  VJL  cos  0P  (7) 

Total  volt-amperes  =  2VPIP  =  3V  p  —  £= 

=  V3VJL  (8) 

where  6P  is  the  phase  power-factor  angle,  i.e.,  the  phase  angle 
between  any  phase  current  and  the  corresponding  phase  voltage. 
For  sinusoidal  waves  and  a  balanced   Y-connected   circuit, 
equations  (5)  and  (6)  become 

Total  power  =  P0  =  3  ^~  /p(p.f  -)P 
y  3 

=  V3VJL  cos  0P  (9) 


316  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Total  volt-amperes  =  SFP7P  =  3  — JL  Ip 

=  V*VJL  (10) 

From  equations  (7),  (8),  (9)  and  (10),  it  is  evident  that  the 
expressions  for  the  total  power  and  the  total  volt-amperes  of  a 
balanced  three-phase  circuit,  having  sinusoidal  waves  of  current 
and  voltage,  are  the  same  whether  the  circuit  is  A- or  Y-connected. 

Equations  (7),  (8),  (9)  and  (10)  also  hold  for  balanced  condi- 
tions when  the  current  and  voltage  waves  are  not  sinusoidal, 
provided  there  are  no  harmonics  present  of  triple  frequency  or 
any  multiple  of  that  frequency.  This  follows  from  the  fact 
that  when  a  balanced  three-phase  circuit  contains  no  harmonics 
of  triple  frequency  or  multiples  of  this  frequency  the  ratio  of  line 
to  phase  voltage  for  wye  connection  and  the  ratio  of  line  to 
phase  current  for  delta  connection  are  equal  to  the  square  root  of 
three.  (See  pages  290  and  294.)  When  equations  (7),  (8),  (9) 
and  (10)  are  applied  to  a  balanced  circuit  having  non-sinusoidal 
waves,  but  no  harmonics  of  triple  frequency  or  any  multiple  of 
this  frequency,  VL  and  IL  must  be  understood  to  be  the  equiva- 
lent sinusoidal  voltage  and  current  respectively.  The  angle  Bp 
is  the  equivalent  phase  angle  for  the  equivalent  sinusoidal  waves 
in  each  phase. 

Since  there  can  be  no  third  harmonics  in  the  line  voltages  or 
line  currents  of  a  balanced  three-phase  system,  there  can  be  no 
third  harmonics  in  the  equivalent  wye  voltages  or  equivalent 
delta  currents  of  such  a  system.  There  are  only  a  few  cases 
where  there  will  be  third  harmonics  in  the  actual  wye  voltages 
of  a  circuit  or  in  the  actual  delta  currents  of  a  circuit,  when 
balanced  line  voltages  are  impressed. 

From  equations  (9)  and  (10)  it  is  obvious  that  the  power-factor 
of  a  balanced  three-phase  circuit,  which  does  not  contain  har- 
monics of  triple  frequency  or  multiples  of  triple  frequency  in 
either  voltage  or  current  in  any  of  its  phases,  is  given  by 

(Pj.)p   =      /~Pv     r  (ID 

V3    VL  IL 

For  sinusoidal  waves  of  voltage  and  current,  this  is  equal  to 
the  cosine  of  the  phase  angle  between  the  phase  current  and 
phase  voltage.  Power-factor  is  never  the  cosine  of  the  phase 
angle  between  the  line  current  and  line  voltage. 


POWER  AND  POWER-FACTOR  317 

The  power-factor  of  a  balanced  four-phase  circuit  is  given  by 
the  following  expression 

x  (p'f')p  -   Po 


For  star  connection  and  balanced  conditions,  the  line  voltage 
is  equal  to  the  phase  voltage  multiplied  by  the  square  root  of 
two,  and  the  line  and  phase  currents  are  equal.  This  statement 
is  independent  of  the  wave  form  of  the  current  or  voltage. 
Therefore  for  a  balanced,  four-phase,  star-connected  circuit 


4   VL_  T         2-s/2FJz,  (13) 

\/2 

The  same  expression  holds  for  a  balanced,  mesh-connected, 
four-phase  circuit.  For  a  balanced,  mesh-connected,  four-phase 
circuit,  the  line  current  is  equal  to  the  phase  current  multiplied 
by  the  square  root  of  two,  and  the  line  and  phase  voltages  are 
the  same.  This  statement  is  independent  of  the  wave  form  of 
the  current  or  voltage.  Therefore,  for  a  balanced,  four-phase, 
mesh-connected  circuit 

(  =        P°       =  -— ^- 

AV     IL        ZVZVJL  (14) 

^  '  L      7^. 

V2 

Equations  (13)  and  (14)  are  true  for  any  balanced  four-phase 
circuit,  whether  star-  or  mesh-connected,  and  are  independent  of 
wave  form.  For  sinusoidal  waves  of  current  and  voltage,  equa- 
tions (13)  and  (14)  reduce  to  cos  6P,  where  Bp  is  the  phase  angle 
between  the  phase  current  and  the  phase  voltage. 

Power-factor  of  an  Unbalanced  Polyphase  Circuit. — The 
power-factor  of  a  balanced  polyphase  circuit  is  a  perfectly 
definite  thing  which  may  be  determined  by  simple  measurements. 
For  a  balanced  polyphase  circuit  the  power-factors  of  all  phases 
are  equal.  A  satisfactory  workable  definition  for  the  power- 
factor  of  an  unbalanced  polyphase  circuit  is  still  to  be  determined. 
The  question  of  the  definition  of  the  power-factor  of  an  un- 
balanced polyphase  circuit  is  under  discussion  at  the  present 
time  (1921)  by  the  Standards  Committee  of  the  American 
Institute  of  Electrical  Engineers. 

The  power-factor  of  an  unbalanced  polyphase  circuit  might  be 


318  PRINCIPLES  OF  ALTERNATING  CURRENTS 

defined  as  the  average  phase  power-factor,  but  if  so  defined  it 
would  be  impossible  of  determination,  except  by  measuring  the 
current,  voltage  and  power  of  each  phase  separately.  This  is 
impracticable  in  the  majority  of  cases  and  often  would  be 
impossible.  For  example,  if  the  average  power-factor  of  an 
unbalanced  A-connected  circuit  were  to  be  determined,  it  would 
be  necessary  to  open  up  the  delta  and  insert  an  ammeter  and  a 
wattmeter  in  each  phase  in  order  to  measure  the  current  and 
power  of  each  phase  separately.  Line  and  phase  voltage  of  a 
A-connected  circuit  are  the  same.  Although  in  experimental 
work  it  might  be  possible  to  open  a  A-connected  circuit  in  order 
to  insert  instruments,  in  commercial  practice  it  would  seldom 
be  possible  to  do  this.  To  determine  the  average  power-factor 
of  a  Y-connected  circuit,  it  would  be  necessary  to  have  the 
common  junction  between  branches  of  the  load  available,  in 
order  to  measure  the  voltage  and  power  of  each  phase.  In 
many  cases  the  neutral  connections  of  Y-connected  circuits  are 
not  brought  out  or  are  not  available  at  the  point  at  which  the 
measurements  have  to  be  made.  If  the  average  power-factor 
of  a  load  on  a  power  station  were  to  be  determined,  the  only 
measurements  that  could  be  made  would  be  the  line  currents, 
the  voltages  between  the  three  pairs  of  lines  or  conductors 
and  the  total  power.  These  measurements  alone  would  not  be 
sufficient  to  determine  the  average  power-factor  of  the  load. 

In  many  cases  equation  (11),  page  316,  is  used  to  determine 
the  power-factor  of  an  unbalanced  three-phase  circuit  by  sub- 
stituting for  VL  and  IL  the  average  voltage  between  lines  and 
the  average  line  current  respectively.  The  result  obtained  in 
this  way  has  no  particular  significance.  It  is  not  the  average 
phase  power-factor  or  the  total  power  divided  by  the  total 
volt-amperes,  or  anything  else  in  particular. 

If  P,  I  and  V  with  subscripts  1,  2  and  3  represent  phase  power, 
phase  current  and  phase  voltage  of  an  unbalanced  three-phase 
circuit,  the  average  phase  power-factor  of  such  a  circuit  is  given 
by  the  following  expression 


3  ( V I         VI        VI- 

Pl(F2/2)(F8/3)     +Pl(FlJl)(1 


(15) 


POWER  AND  POWER-FACTOR  319 

The  expression  derived  from  equation  (11)  by  substituting 
average  line  current  for  IL  and  average  voltage  between  lines 
for  VL  is 

(    n    =  Pi  +  P*  +  P* 


(16) 


where  Viz,  V&  and  V3i  are  the  voltages  between  the  pairs  of 
lines  1-2,  2-3  and  3-1  respectively,  and  ILi,  ILZ  and  ILS  are  the 
three  line  currents. 

It  is  obvious,  by  comparing  equations  (15)  and  (16),  that 
equation  (16)  can  give  the  average  power-factor  only  when  the 
load  is  balanced,  i.e.,  when  Viz  =  Vzs  =  FSI  and  IL\  =  ILZ  =  /«. 

For  small  amounts  of  unbalancing,  the  average  power-factor 
and  the  value  of  the  power-factor  calculated  from  equation  (16) 
do  not  differ  greatly  in  most  cases,  but  when  equation  (16)  is 
used,  the  fact  should  not  be  overlooked  that  it  does  not  give  the 
average  phase  power-factor.  Commercial  circuits  as  a  rule  are 
not  much  out  of  balance. 

It  has  been  suggested  (1924),  by  a  special  committee  appointed 
for  the  purpose  of  denning  the  power-factor  of  a  polyphase 
circuit,  that  the  power-factor  of  such  a  circuit  be  defined  as  the 
ratio  of  the  total  active  power  to  the  square  root  of  the  sum  of  the 
squares  of  the  total  active  power  and  the  total  reactive  power, 
i.e., 

,    -, Total  active  power 

V  (Total  active  power)2  +  (Total  reactive  power)2 
This  definition  does  not  give  the  ratio  of  watts  to  volt-amperes 
except  when  the  circuit  is  free  from  harmonics. 

The  total  active  power  of  any  balanced  or  unbalanced  three- 
phase  circuit,  that  does  not  have  a  neutral  connection  which 
carries  current,  may  be  measured  by  the  two-wattmeter  method 
for  measuring  power  or  by  a  polyphase  wattmeter  (page  326). 
The  total  reactive  power  of  a  balanced  or  unbalanced  three-phase 
circuit,  that  does  not  have  a  neutral  connection  which  carries 
current,  may  be  measured  by  two  reactive-power  meters  (pages 
63  and  327a)  or  by  a  polyphase  reactive-power  meter  provided 
the  current  and  voltage  are  sinusoidal. 

If  power-factor  is  defined  in  the  way  suggested  by  the  com- 
mittee, it  is  possible  to  determine  the  power-factor  of  a  balanced 


320  PRINCIPLES  OF  ALTERNATING  CURRENTS 

or  an  unbalanced  three-phase  circuit  as  easily  as  the  power-factor 
of  a  single-phase  circuit.  Since  the  maximum  output  of  all 
alternating-current  generators,  motors  and  transformers  and  the 
regulation  and  capacity  of  transmission  and  distribution  lines 
depend  not  only  upon  the  amount  of  power  developed  but  also 
upon  the  power-factor  at  which  they  operate,  it  is  extremely 
important  to  have  a  definition  of  the  power-factor  of  a  polyphase 
circuit  which  makes  it  possible  to  determine  the  power-factor 
of  such  a  circuit  easily.  Power-factor  may  then  be  taken  into 
consideration  in  determining  equitable  power  rates  for  customers 
who  use  power  at  relatively  low  power-factor. 

Relative  Amounts  of  Copper  Required  to  Transmit  a  Given 
Amount  of  Power  a  Fixed  Distance,  with  a  Fixed  Line  Loss  and 
Fixed  Voltage  between  Conductors,  over  a  Three-phase  Trans- 
mission Line  under  Balanced  Conditions  and  over  a  Single-phase 
Line.  —  Let  P  and  V  be  respectively  the  power  transmitted  and 
the  limiting  voltage  between  conductors.  Let  I\  and  73  be  the 
currents  per  conductor  for  the  single-phase  and  three-phase 
transmission  respectively.  Then 

P(single-phase)  =  VI  \  cos  6P  (17) 

P(three-phase)  =  V3F/3  cos  6P  (18) 

where  Bp  is  the  power-factor  angle,  i.e.,  the  angle  between  the 
line  current  and  line  voltage  for  single-phase  transmission  and 
the  angle  between  the  line  current  and  the  equivalent  wye 
voltage  for  the  three-phase  transmission.  Balanced  three-phase 
transmission  is  assumed. 

From  equations  (17)  and  (18) 


Let  ri  and  r3  be  the  resistance  per  conductor  of  the  single-phase 
and    three-phase    lines    respectively.     Then   for   equal    copper 
losses  for  the  single-phase  and  three-phase  transmission 
27,  V,  =  373V3 

'J  _  3       73*  _  3       1  __  1 
r3~2X7?~2X3~2 

Since  the  amount  of  copper  required  for  a  given  length  of 
transmission   line   is   directly   proportional   to   the   number    of 


POWER  AND  POWER-FACTOR  321 

conductors  employed  and  inversely  proportional  to  the  resistance 
per  conductor 

Copper  for  three-phase  transmission  _  3       1  _  3 
Copper  for  single-phase  transmission       2       2  ~~  4 
From  equation  (21)  it  is  evident  that  twenty-five  per  cent  less 
copper  is  required  to  transmit  a  given  amount  of  power  a  given 
distance,  with  a  fixed  transmission  loss  (copper  loss)  and  a   fixed 
voltage    between    conductors,    over   a   three-phase   line    under 
balanced  conditions  than  over  a  single-phase  line,  or  thirty-three 
and  a  third  per  cent  more  copper  is  required  for  the  single-phase 
transmission. 

Relative  Amounts  of  Copper  Required  to  Transmit  a  Given 
Amount  of  Power  a  Fixed  Distance,  with  a  Fixed  Line  Loss  and  a 
Fixed  Voltage  between  Conductors,  over  a  Four -phase  Trans- 
mission Line  under  Balanced  Conditions  and  over  a  Single-phase 
Line. — The  maximum  voltage  between  conductors  of  a  four-phase 
line  exists  between  alternate  conductors.  If  V  represents  this 
voltage,  the  voltage  between  adjacent  conductors  is 

A/2 

-~  X  V  =  0.7077 

If  the  limiting  voltage,  V,  is  taken  as  the  maximum  voltage 
between  any  two  of  the  conductors  of  the  four-phase  line,  the 
four-phase  system  will  require  the  same  amount  of  copper  as  a 
single-phase  line  for  the  transmission  of  a  given  amount  of 
power,  under  balanced  conditions,  a  fixed  distance  with  a  fixed 
line  loss.  The  four-phase  system  is  equivalent  to  two  single- 
phase  lines  each  transmitting  half  the  total  power  at  a  voltage 
equal  to  the  diametrical  voltage  of  the  system.  Twice  as  many 
conductors  are  required  as  for  single-phase  transmission,  but 
each  conductor  need  be  only  half  as  large,  as  it  carries  only  half 
as  much  current  as  each  conductor  of  the  single-phase  line. 

If  the  voltage  between  adjacent  conductors  of  the  four-phase 
line  is  made  equal  to  the  voltage,  7,  between  conductors  of  the 
single-phase  line,  the  maximum  voltage  between  any  two  con- 
ductors of  the  four-phase  line  will  be  \/2  X  V  and  will  occur 
between  alternate  conductors. 

If  V  is  taken  as  the  voltage  between  adjacent  conductors  of 
the  four-phase  line  and  74  is  used  as  the  current  per  conductor 


322  PRINCIPLES  OF  ALTERNATING  CURRENTS 

y 

P  (four-phase)  =  4  — 7=  I*  cos  Op 

V  2 

=  2A/27/4  cos  0P  (22) 

P  (single-phase)  =  71 1  cos  Bp  (23) 

Since  the  power  is  the  same  in  the  two  cases 

i4  =  — ^7=  (24) 

Ii       2V2 

If  r4  is  the  resistance  per  conductor  of  the  four-phase  line, 
for  equal  losses  for  single-phase  and  four-phase  transmission 
2/i2ri  =  4  /4V4 

n  =  2  x  ^!  =  2  x  /    1    N2 
r4  /i2 


Copper  for  four-phase  transmission     __  4       1.  =  1.      .     , 
Copper  for  single-phase  transmission  ~  2       4       2 
Relative  Amounts  of  Copper  Required  to  Transmit  a  Given 
Amount  of  Power  a  Fixed  Distance,  with  a  Fixed  Line  Loss  and  a 
Fixed  Voltage  to  Neutral,  when  the  Loads  are  Balanced. — When 
the  voltage  to  neutral  is  fixed,  there  is  no  difference  in  the  amounts 
of  copper  required  to  transmit  a  given  amount  of  power  a  fixed 
distance  with  a  fixed  line  loss,  whether  the  transmission  be  single- 
phase,   three-phase  or  four-phase.     Let   Vn  be  the  voltage  to 
neutral.     Then 

P(single-phase)  =  2VnI1  cos  6P  (27) 

P(three-phase)  =  3VaI3  cos  6P  (28) 

P  (four-phase)    =  47n/4  cos  0P  (29) 

Is  _  2  /4  =  2 

Ii  ~  3  /i       4 

2Ii*ri  =  3I32r3 

ri_3l3!_34_2 

r3  ~  2  X  /!2  ~  2  X  9  "  3 
2/iV!  =  4742r4 

r1_4742=44_l 

r4       2  A  7i2       2  A  16       2 

Copper  for  three-phase  transmission  __  3       2  __  ^ 
Copper  for  single-phase  transmission       2       3 
Copper  for  four-phase  transmission     _  4       1  _  ^ 
Copper  for  single-phase  transmission       2       2 


POWER  AND  POWER-FACTOR 


323 


Power  Measurements  in  Three-phase  Circuits. — The  total 
power  in  any  polyphase  circuit  is  equal  to  the  algebraic  sum  of 
the  powers  in  each  phase.  If  a  wattmeter  is  placed  in  each 
phase,  the  sum  of  the  wattmeter  readings  is  the  true  power  in 
the  circuit.  Consider  the  three-phase  A-connected  and  Y- 
connected  circuits  shown  in  Fig.  96.  The  circles  marked  W  in 
this  figure  represent  the  current  coils  of  the  wattmeters.  The 
rectangles  in  the  middle  of  the  circles  are  the  potential  coils. 

The  current  coil  of  each  wattmeter  carries  the  current  in  one 
phase.  The  voltage  of  this  phase  is  impressed  across  the  poten- 
tial coil  of  the  wattmeter.  Obviously  each  wattmeter,  connected 
as  shown  in  Fig.  96,  measures  the  power  in  a  single  phase.  The 


FIG.  96. 

sum  of  the  readings  of  the  three  wattmeters,  for  either  the  delta 
connection  or  the  wye  connection,  must  give  the  total  power  in 
the  circuit. 

Under  ordinary  conditions,  it  is  impossible  to  break  into  a 
A-connected  circuit  and  thus  place  a  wattmeter  in  each  phase  as 
shown  in  the  left-hand  half  of  Fig.  96.  Neither  is  it  always  pos- 
sible, in  the  case  of  a  Y-connected  circuit,  to  get  at  the  neutral 
point  which  is  required  for  the  connections  shown  in  the  right- 
hand  half  of  Fig.  96.  If  the  connection  shown  in  the  right-hand 
diagram,  between  the  common  junction  of  the  wattmeter  poten- 
tial coils  and  the  neutral  point  of  the  circuit,  is  omitted,  the  sum 
of  the  readings  of  the  three  wattmeters — really  the  algebraic 
sum — will  give  the  true  power  in  the  circuit  for  any  degree  of 
unbalancing  or  for  any  wave  form,  whether  the  circuit  be  A- 
connected  or  Y-connected,  provided  the  load,  if  Y-connected, 
does  not  have  its  neutral  point  connected  to  the  neutral  of  the 


324 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


FIG.  97. 


source  of  power.  If  the  neutrals  are  interconnected,  but  the 
neutral  connection  does  not  carry  current,  the  conditions  are  the 
same  as  if  it  did  not  exist.  Even  with  a  balanced  load,  the  neutral 
connection,  if  it  exists,  will  carry  current  if  the  phase  voltage  of  the 
circuit  contains  a  harmonic  of  triple  frequency  or  any  multiple  of 
triple  frequency.  When  there  is  no  harmonic  of  triple  frequency 
or  multiple  of  this  frequency  in  the  phase  voltage  of  the  source 
of  power,  there  will  still  be  a  pronounced  third-harmonic  current 

in  the  neutral  connection  under  bal- 
anced conditions,  when  the  load  is 
inductive  and  contains  iron,  especially 
if  the  iron  is  worked  at  high  saturation. 
Proof  of  the  Three -wattmeter 
Method  for  Measuring  the  Power  in 
a  Three-phase  Circuit. — Let  the 
instantaneous  values  of  the  phase  cur- 
rents, phase  voltages  and  phase 
powers  be  denoted  by  i,  e  and  p, 
respectively,  with  subscripts  1,  2  and 

3  to  indicate  the  particular  phase.  Let  the  three  wattmeters 
Be  connected  as  shown  in  Fig.  97. 

The  instantaneous  power  in  any  single-phase  circuit  is  equal 
to  the  product  of  the  instantaneous  values  of  current  and  voltage. 
Therefore  the  total  instantaneous  power,  po,  in  the  three-phase 
circuit  is 

Po  =  ioieoi  +  ^02^02  +  iotfoa  (32) 

According  to  Kirchhoff's  laws 

ioi  +  ioz  +  ioa  =  0  (33) 

Let  e0 a  be  the  instantaneous  potential  difference  between  the 
points  0  and  a,  Fig.  97,  i.e.,  the  difference  in  potential  between 
the  neutral  point  of  the  system  and  the  common  junction  of  the 
potential  coils  of  the  wattmeters.  Then  the  following  relation 
is  obviously  true. 

eoa(ioi  -f  io2  +  ios)  =  0 

eoaioi  +  e0aioz  +  e0aioa  =  0  (34) 

Subtracting  equation  (34)  from  equation  (32)  gives 


POWER  AND  POWER-FACTOR  325 

oa)  +  ^02(^02  —  eoa)  H-  ias(eoz  —  e0a)  =  PQ      (35) 


But 


Therefore,  the  instantaneous  power  in  the  circuit  is 

PQ    =   ioi€ai   +  iotCaZ   +  ^03^3  (36) 

=    Pi   +   P2   +   P3  (37) 

where  pi,  p2  and  p3  are  the  instantaneous  powers  corresponding 
to  ioiBai,  iotfaz  and  tosias  respectively.  The  average  power  in  the 
circuit  is 


I    f 

Tjo 


(38) 


which  is  the  sum  of  the  actual  wattmeter  readings. 

This  method  of  measuring  power  in  a  three-phase  circuit  is 
known  as  the  three-  wattmeter  method.  Since  any  A-connected 
circuit  may  always  be  replaced  by  an  equivalent  Y-connected 
circuit,  it  is  evident  that  the  three-wattmeter  method  of  measur- 
ing power  may  be  used  to  measure  the  total  power  in  a  A-con- 
nected circuit  as  well  as  to  measure  the  total  power  in  a 
Y-connected  circuit. 

The  proof  of  the  three-wattmeter  method  is  based  on  the 
assumption  that  the  sum  of  the  instantaneous  currents  in  the 
phases  of  the  load  is  zero.  If  there  is  a  neutral  connection 
between  the  load  and  the  source  of  power,  and  it  carries  current,  the 
sum  of  the  three  instantaneous  phase  currents  is  no  longer  equal 
to  zero.  When  the  neutrals  of  the  load  and  source  of  power  are 
connected  and  the  neutral  connection  carries  a  current,  io», 
equation  (33)  becomes 

ion  +  1*01  +  ioz  +  ios  =  0 

The  sum  of  the  currents  ioi,  i0z  and  io3  is  no  longer  zero,  except 
when  the  neutral  current  is  zero.  The  proof  of  the  three-watt- 
meter method  for  measuring  power  holds  only  when  there  is  no 
neutral  connection  between  the  source  of  power  and  the  load 
or  when  the  neutral  connection,  if  used,  carries  no  current.  If 


326  PRINCIPLES  OF  ALTERNATING  CURRENTS 

the  neutral  point,  a,  of  the  wattmeters  is  connected  to  the  neu- 
tral point,  0,  of  the  load  (Fig.  97,  page  324),  the  three- wattmeter 
method  of  measuring  power  will  give  the  true  power  under  all 
conditions,  since  there  is  a  wattmeter  in  each  phase  of  the  load. 
The  N-wattmeter  Method  for  Measuring  the  Power  in  an 
N -phase  Circuit. — The  proof  just  given  is  not  limited  to  three- 
phase  circuits.  It  may  be  extended  to  apply  to  a  circuit  with 
any  number  of  phases.  The  power  in  an  n-phase  circuit  may  be 
measured  by  n  wattmeters,  each  with  its  current  coil  in  one  line 
and  with  its  potential  coil  connected  between  the  line  in  which 
its  current  coil  is  placed  and  a  common  junction  point,  to  which 
one  terminal  of  each  wattmeter  is  connected.  The  n-wattmeter 
method  does  not  hold  when  there  is  a  neutral  connection,  which 
carries  current,  between  the  load  and  the  source  of  power,  except 
when  the  common  junction  of  the  potential  coils  of  the  n  watt- 
meters is  connected  to  the  neutral  of  the  load.  The  three-watt- 
meter method  for  measuring  three-phase  power  or,  in  general, 
the  n-wattmeter  method  for  measuring  n-phase  power  is  seldom 
used.  Three-phase  power  is  usually  measured  by  the  two- watt- 
meter method.  In  general  the  (n  —  l)-wattmeter  method  is 
used  for  measuring  the  power  in  an  n-phase  circuit. 

Two -wattmeter  Method  for  Measuring  Power  in  a  Three- 
phase  Circuit  and  the  (N— 1) -wattmeter  Method  for  Measuring 
Power  in  an  N-phase  Circuit. — No  assumption  was  made  in  the 
proof  of  the  three-wattmeter  method  regarding  the  position  of 

the  common  junction  point,  a,  of  the 
"wattmeter  potential  coils,  which  may 
be  any  point  whatsoever.  It  may  be 
on  one  of  the  lines.  In  this  case,  one 
wattmeter  will  read  zero  and  may  be 
omitted.  The  algebraic  sum  of  the 
readings  of  the  two  remaining  watt- 
meters will  then  give  the  true  power. 
If  there  are  more  than  three  phases, 
the  common  junction  point  of  the 

potential  coils  of  the  n  wattmeters  may  still  be  on  one  of  the 
lines  making  the  reading  of  one  of  the  n  wattmeters  zero.  The 
algebraic  sum  of  the  readings  of  the  other  (n  —  1)  wattmeters 
will  be  equal  to  the  true  power  in  the  circuit. 


POWER  AND  POWER-FACTOR  327 

In  general,  the  power  in  an  n-phase  circuit  may  be  measured  by 
(n  —  1)  wattmeters,  each  with  its  current  coil  in  one  line  and  its 
potential  coil  bridged  between  the  line  containing  its  current  coil 
and  the  line  which  does  not  contain  the  current  coil  of  a  watt- 
meter. Applied  to  a  three-phase  circuit  the  (n  —  1) -wattmeter 
method  becomes  the  two- wattmeter  method.  The  connections 
for  the  two-wattmeter  method  for  measuring  power  in  a  three- 
phase  circuit  are  shown  in  Fig.  98,  page  326. 

1  CT  I  CT 

^°  =  T  i    i01  e*ldt  ~^  T  I    * 

The  two-wattmeter  method,  or  in  general,  the  (n  —  1)- 
wattmeter  method,  gives  the  true  power  in  a  circuit  without 
regard  to  balance  or  wave  form,  provided  the  neutral  of  the  load, 
if  star-connected,  does  not  have  its  neutral  connected  to  the  neu- 
tral of  the  source  of  power.  If  the  neutral  connection  carries  no 
current,  the  conditions  are  the  same  as  if  the  neutral  connection 
did  not  exist.  About  the  only  conditions  likely  to  occur  in 
practice,  when  a  neutral  connection  for  a  three-phase  circuit  will 
not  carry  current,  are  when  the  load  is  exactly  balanced  and 
there  are  no  harmonics  present  of  triple  frequency  or  any  multiple 
of  that  frequency. 

The  connections  of  the  wattmeters  for  the  three- wattmeter 
method  are  symmetrical.  The  readings  of  the  three  wattmeters, 
for  the  three-wattmeter  method  of  measuring  the  power  in  a 
three-phase  circuit  therefore,  will  either  be  all  positive  or  all  nega- 
tive, barring  very  exceptional  conditions,  and  will  add  directly  to 
give  the  true  power  in  the  circuit.  The  only  case  where  the  read- 
ings of  the  wattmeters  will  not  either  be  all  positive  or  all  negative 
is  when  the  load  is  very  badly  unbalanced.  Under  all  conditions, 
the  algebraic  sum  of  the  wattmeter  readings  is  the  true 
power. 

With  the  two-wattmeter  method,  the  wattmeters  are  not 
connected  symmetrically.  By  referring  to  Fig.  98,  it  will  be 
seen  that  wattmeter  No.  1  has  its  potential  coil  connected  from 
line  1  to  line  3  or  in  a  left-hand  direction  with  respect  to  the 
sequence  of  phases  as  numbered.  Wattmeter  No.  2  has  its 
potential  coil  connected  from  line  2  to  line  3  or  in  a  right-hand 
direction.  The  algebraic  sum,  not  the  numerical  sum,  of  the 


327 'a  PRINCIPLES  OF  ALTERNATING  CURRENT 

readings  of  the  two  wattmeters  always  gives  the  true  power. 
In  order  to  read  a  wattmeter  it  must  be  connected  so  that  its 
pointer  will  deflect  up  scale.  It  is  necessary,  therefore,  to  so 
connect  the  two  wattmeters  that  their  pointers  will  deflect  up 
scale  and  then  to  determine,  from  the  connections  used,  whether 
the  readings  of  the  wattmeters  are  alike  in  sign  or  opposite. 
For  balanced  loads,  the  readings  will  be  opposite  in  sign  whenever 
the  power-factor  of  the  circuit  is  less  than  five-tenths. 

To  determine  whether  the  readings  of  the  wattmeters  used  in 
the  two-wattmeter  method  for  measuring  power  in  a  three-phase 
circuit  are  of  like  or  unlike  sign,  it  is  merely  necessary  to  see 
whether  the  wattmeters  are  connected  alike  or  differently,  i.e.,  to 
see  whether  the  current  is  lead  to  corresponding  current  terminals 
and  whether  corresponding  ends  of  the  potential  coils  are  con- 
nected to  the  common  line.  If  the  wattmeters  are  of  different 
make  or  type  they  may  both  be  placed  in  the  same  circuit  to 
determine  which  terminals  correspond.  If  the  load  is  approxi- 
mately balanced,  the  sign  of  the  readings  of  the  wattmeters  may 
be  determined  very  easily  by  merely  disconnecting  from  the 
common  line  (line  3  in  Fig.  98)  the  potential  coil  of  the  watt- 
meter which  has  the  smaller  deflection  and  connecting  it  to  the 
line  not  containing  its  current  coil.  If  the  reading  of  the  watt- 
meter reverses  when  this  is  done,  the  signs  of  the  wattmeter 
readings  are  opposite  and  the  readings  must  be  subtracted  to  give 
the  true  power.  Changing  the  connections  of  the  wattmeter  with 
the  smaller  deflection,  as  indicated,  connects  the  potential  coils  of 
both  wattmeters  alike  with  respect  to  the  cyclic  order  of  the 
three-phase  circuit  and  serves  the  same  purpose  as  placing  both 
wattmeters  in  the  same  circuit. 

The  Total  Reactive  Power  of  a  Three-phase  Circuit,  whose 
Wave  Forms  are  Sinusoidal  and  which  has  no  Neutral  Connec- 
tion that  Carries  Current,  may  be  Measured  by  the  Use  of  Two 
Reactive-power  Meters  Connected  like  the  Wattmeters  for 
Measuring  Power  by  the  Two-wattmeter  Method  for  Measuring 
Power  in  a  Three-phase  Circuit.— Let  the  current  and  voltage 
waves  of  a  three-phase  circuit  be  sinusoidal.  They  may  then  be 
represented  by  vectors  each  of  which  may  be  resolved  in  to  a 
real  and  an  imaginary  component.  Let  small  v's  with  and  with- 


PO  WER  AND  POWER-FACTOR  3276 

out  primes  represent  respectively  imaginary  and  real  components 
of  the  voltages.     Let  small  i's  with  and  without  primes  represent 
respectively  corresponding  components  of  the  currents.     Assume 
that  there  is  no  neutral  current. 
Refer  to  Fig.  98,  page  326.     Let 

7oi  =  ii  +  jii 
7  02  =  it  +  jit' 

7os  =  is  +  jis 
Also  let 

Voi   =   Vi  +  jVi 

702    = 


Then,  referring  to  page  66,  the  total  reactive  power  is 

Pr  =  Wii  -  wiii')  +  Wit  -  vtit')  +  Wit  ~  fW)   (39a) 
Since  there  is  assumed  to  be  no  neutral  current, 

1  01   +  /02    +   /03    =    0 

The  following  relations  must  also  be  true. 

ii  +  12  +  *s  =  0  (396) 

ii'  +  U  +  i*    =  0  (39c) 

Substituting  the  values  of  i3  and  isf  from  equations   (396) 
and  (39c)  in  equation  (39a)  gives 

Pr  =  vi'i\  —  viii  -h  v*i*  —  v2i2   —  v3'(ii  +  it)  +  va(ii   +  it) 
=  iiW  -  vs)  +  ii(-Vi  +  v3)  -f  12^2'  -  Vs')  H-  i2(-vz  +v3) 
-  iiv^i)  +  (^2^32'  -  ^2^32)  (39d) 


where  P\r  and  P2r  are  respectively  the  readings  of  reactive-power 
meters  connected  as  shown  in  Fig.  98,  page  326. 


328  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Problem  Illustrating  the  Use  of  the  Two-wattmeter  Method 
for  Measuring  Power  in  a  Balanced  Three-phase  Circuit. — Two 

wattmeters  are  connected  to  measure  the  power  taken  by  a 
certain  balanced,  three-phase  inductive  load.  Wattmeter  No.  1 
has  its  current  coil  in  line  1  and  its  potental  coil  between  lines  1 
and  3.  Wattmeter  No.  2  has  its  current  coil  in  line  2  and  its 
potential  coil  between  lines  2  and  3.  The  readings  of  the  watt- 
meters are  respectively  40.96  and  4.36  kilowatts.  When  that 
end  of  the  potential  coil  of  wattmeter  No.  2  which  is  connected 
to  line  3  is  connected  to  line  1,  the  reading  of  wattmeter  No.  2 
reverses  and  the  pointer  goes  up  against  the  stop.  If  the  con- 
nection of  the  current  coil  of  this  wattmeter  is  now  reversed, 
so  that  the  pointer  will  deflect  up  scale,  the  wattmeter  reads  40.96 
kilowatts. 

(a)  What  is  the  power  taken  by  the  load? 

(6)  If  the  line  currents  are  each  100  amperes  and  the  voltages 
between  lines  are  each  500  volts,  what  is  the  power-factor  of  the 
load? 

(c)  By  what  angle  does  the  equivalent  sine  current  in  each 
phase  lag  the  equivalent  sine  voltage  impressed  across  each 
phase? 

(d)  If  the  load  is  Y-connected  without  a  neutral,  what  is  the 
current  in  each  phase  and  what  is  the  voltage  across  the  terminals 
of  each  phase? 

(e)  If  the  load  had  been  A-connected,  what  would  have  been 
the  current  in  each  phase  and  what  would  have  been  the  voltage 
across  the  terminals  of  each  phase? 

Since  there  is  no  neutral  connection,  when  the  load  is  con- 
nected in  wye  there  can  be  no  harmonic  of  triple  frequency  or 
multiple  of  this  frequency  in  the  current.  There  can  be  no 
harmonic  of  triple  frequency  or  multiple  of  this  frequency  in  the 
line  voltage,  since  the  load  is  balanced.  There  would  be  a  third 
harmonic  in  the  phase  voltage,  i.e.,  the  wye  voltage,  if  the  load 
involved  iron.  It  will  be  assumed  that  no  iron  is  present.  In 
this  case  there  can  be  no  harmonic  of  triple  frequency  or  multiple 
of  this  frequency  in  any  part  of  the  circuit.  Under  this  condi- 
tion, the  ratio  of  the  line  and  phase  (wye)  voltages  will  be  equal 
to  the  square  root  of  three.  If  no  iron  is  present,  there  will  be  no 
circulatory  current  of  triple  frequency  or  any  multiple  of  this 


POWER  AND  POWER-FACTOR  329 

frequency  in  the  closed  delta,  when  delta  connection  is  assumed. 
Under  this  condition,  the  ratio  of  line  and  phase  (delta)  currents 
will  be  equal  to  the  square  root  of  three. 

(a) 

Since  wattmeter  No.  2  reverses  when  the  connections  of  its 
potential  coil  are  changed,  its  reading  must  have  been  negative 
as  originally  connected.  Therefore, 

Total  power  =  F0  =  40.96  -  4.36  =  36.60  kilowatts. 


Po  36.60 

= 


V3  /«„.  Vline    -  V3X100X500 
=  0.4226 

(c) 

Angle  of  lag  of  the  equivalent  sine  phase  current  behind  the  equi- 
valent sine  phase  voltage  for  either  wye  or  delta  connection  is 

6P  =  cos-1  0.4226  =  65  degrees. 

(d) 

=  100  amperes. 


Wye  connection 


500 


=  288.7  volts. 
V'6 

w 

7. 

Delta  connection 


100 

phase  =  ~=  =  57.7  amperes. 


V phase  =    500  VOltS. 

Another  Example  of  the  Use  of  the  Two-wattmeter  Method.— 

It  is  frequently  convenient,  when  solving  problems  in  three-phase 
circuits,  to  apply  the  principle  of  the  two-wattmeter  method  to 
determine  the  total  power  taken  by  a  circuit.  When  the  solution 
of  a  problem  gives  the  line  currents  and  line  voltages  expressed  in 
their  complex  form,  the  easiest  way  to  determine  the  total  power 
consumed  is  to  assume  that  wattmeters  have  been  inserted  in  the 
circuit  to  measure  the  power  by  the  two-wattmeter  method. 


330  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  solution  of  the  problem  illustrating  KirchhofTs  laws, 
which  was  given  on  page  272,  gives  for  the  three  line  currents 

lao  =  20.10  -  J10.44 
ho  =  -10.62  +  jl.83 
Ico  =  -9.48  +  J8.60 

The  line  voltages  are 

F«6  =  230  +  jO 

Vbc  =  -H5  -  J199.2 

yca  =  -115  +  199.2 

Assume  that  one  wattmeter  is  placed  in  line  a  (see  Fig.  80, 
page  272),  with  its  potential  coil  connected  between  lines  a  and  b, 
and  that  the  other  wattmeter  is  placed  in  line  c,  with  its  potential 
coil  connected  between  lines  c  and  6.  The  algebraic  sum  of  the 
readings  of  the  wattmeters,  if  actually  connected  as  indicated, 
would  be  the  total  power  in  the  circuit. 

Wattmeter  in  line  a  would  carry  a  current  Iao  =  20.10  —  jlO.44 
and  would  have  a  voltage  Vab  =  230  +  JO  impressed  across  its 
potential  coil.  This  wattmeter  would  read 

Pa  =  (20.10)(230)  +'(-10.44X0) 
=  4623  +  0  =  4623  watts. 

The  other  wattmeter  would  carry  a  current  Ico  =  —9.48  + 
J8.60  and  would  have  a  voltage  Vcb  =  -Vbc  =  115  +  J199.2 
impressed  across  its  potential  coil.  This  wattmeter  would  read 

pc  =  (_9.48)(115)  +  (8.60)(199.2) 
=  -1090  +  1713  =  623  watts. 
Total  power  P0  =  Pa  +  Pc  =  4623  +  623 

=  5246  watts. 

Relative  Readings  on  Balanced  Loads  of  Wattmeters  Con- 
nected for  the  Two-wattmeter  Method  of  Measuring  Power  in 
a  Three-phase  Circuit  when  the  Current  and  Voltage  Waves 
are  Sinusoidal. — Figure  99  shows  the  vector  diagram  of  a 
balanced,  three-phase,  inductive  load  having  a  power-factor  of 
cos  6P.  For  an  inductive  load  having  a  power-factor  of  cos  8P, 
the  phase  currents  lag  the  phase  voltages  by  an  angle  Bp.  70i 
lags  Foi  by  Bp  degrees,  J02  lags  Voz  by  Qp  degrees  and  Im  lags 
^03  by  Op  degrees. 


POWER  AND  POWER-FACTOR 


331 


Let  the  wattmeters  be  connected  as  shown  in  Fig.  98,  page  326. 
The  wattmeter  in  line  1  carries  the  current  70i  and  has  the  voltage 
Vsi  impressed  across  the  terminals  of  its  potential  coil.  The 
voltage  and  current  are  considered  in  the  same  direction  in 
phase  01.  The  wattmeter  in  line  2  carries  the  current  702  and 
has  the  voltage  V32  impressed  across  the  terminals  of  its  poten- 
tial coil.  The  current  and  voltage  are  considered  in  the  same 


FIG.  99. 


direction  in  phase  02.  /Oi  lags  F3i  by  (0P°  •-  30°)  degrees.  702 
lags  F32  by  (0P°  +  30°)  degrees.  The  readings  of  the  two  watt- 
meters are  therefore 


PiCwattmeter  in  line  1)  =  I01V31  cos  (0P°  -  30°) 

=  ILVL  cos  (0P°  -  30°) 


P2(wattmeter  in  line  2)  = 


cos  (0P°  +  30°) 


=  ILVL  cos  (0P°  +  30°) 


(40) 
(41) 


where  the  subscript  L  indicates  line  values. 

When  0P  =  0,  i.e.,  when  the  power-factor  of  the  circuit  is 
unity,  both  wattmeters  read  alike,  namely  IL\TL  cos  30°. 

When  0P  =  60  degrees,  i.e.,  when  the  power-factor  is  0.5, 
the  current  and  voltage  for  the  wattmeter  in  line  1  will  be  out  of 


332 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


phase  by  30  degrees,  the  current  lagging  the  voltage,  while  the 
current  and  voltage  for  the  wattmeter  in  line  2  will  be  out  of 
phase  by  90  degrees,  the  current  also  lagging.  The  reading  of 
the  wattmeter  in  line  2  will  therefore  be  zero.  The  entire  power 
developed  in  the  circuit  will  be  indicated  by  the  wattmeter  in 
line  1.  For  angles  of  lag  greater  than  sixty  degrees,  i.e.,  for 
power-factors  less  than  0.5,  the  current  and  voltage  for  the  watt- 
meter in  line  2  will  be  out  of  phase  by  more  than  ninety  degrees. 
Under  this  condition,  the  reading  of  the  wattmeter  in  line  2  will 


1.0 
0.8 
0.6 
0.4 
0.2 
0.0 
-0.2 
-0.4 
-0.6 
-0.8 
-1.0 

/ 

z 

/ 

0         7 

0         8 

0        1 

0        2 

0         3 

0         4 

0         5 

o/xe 

D         9 

0        1( 

Powe 
/ 

r-fae6 

rinPe 

r  cent 

/ 

/ 

/ 

/ 

/ 

/ 

Ord 

nates 

.  cos  (0P+300) 

coB(0£-«n 

FIG.   100. 


reverse,  i.e.,  it  will  become  negative,  and  must  be  subtracted 
from  the  reading  of  the  wattmeter  in  line  1  to  get  the  true 
power.  The  true  power  is  always  equal  to  the  algebraic  sum 
of  the  readings  of  the  two  wattmeters,  but  the  sign  of  the  reading 
of  one  of  the  wattmeters  reverses  and  becomes  negative  when 
the  power-factor  of  the  circuit  becomes  less  than  0.5. 

When  the  power-factor  is  less  than  0.5,  it  is  necessary  to  reverse 
the  connections  of  the  current  coil  of  one  of  the  wattmeters  in 


POWER  AND  POWER-FACTOR  333 

order  to  make  it  read  up-scale.  All  readings  taken  after  the 
reversal  of  the  current  coil  must  be  considered  negative. 

The  ratio  of  the  readings  of  the  two  wattmeters,  connected  for 
the  two- wattmeter  method,  is  the  same  for  equal  power-factors 
with  leading  and  lagging  currents,  but  the  actual  readings  are  in- 
terchanged. For  example,  the  wattmeter  in  line  2  will  read  zero 
for  a  power-factor  of  cos  (+60°),  i.e.,  a  lag  of  the  phase  current 
behind  the  phase  voltage  of  60  degrees.  For  an  angle  of  lead 
of  60  degrees,  i.e.,  a  power-factor  of  cos  (  —  60°),  the  wattmeter 
in  line  1  will  read  zero.*  Both  wattmeters  will  read  alike  for 
unity  power-factor.  They  will  read  alike  in  magnitude  but 
opposite  in  sign  for  zero  power-factor.  It  must  be  remembered 
that  the  above  statements  are  true  only  when  the  load  is  balanced 
and  the  current  and  voltage  waves  are  sinusoidal. 

Clockwise  phase  order  is  assumed  in  the  preceding  discus- 
sion. Changing  the  phase  order  will  interchange  the  readings 
of  the  wattmeters. 

The  ratios  of  the  readings  of  two  wattmeters,  connected  to 
measure  the  power  taken  by  a  balanced,  three-phase  load  with 
sinusoidal  current  and  voltage  waves,  are  plotted  against  power- 
factors  in  Fig.  100,  page  332. 

Determination  of  the  Power-factor  of  a  Balanced  Three- 
phase  Circuit,  when  the  Current  and  Voltage  Waves  are  Sinu- 
soidal, from  the  Readings  of  Two  Wattmeters  Connected  to 
Measure  the  Total  Power  by  the  Two-wattmeter  Method. — The 
power-factor  of  a  balanced  three-phase  circuit  may  be  determined 
by  measuring  the  total  power,  the  line  current  and  the  line 
voltage,  and  then  applying  equation  (11),  page  316.  If  a  circuit 
is  balanced  and  its  current  and  voltage  waves  are  sinusoidal, 
the  power-factor  may  be  found  from  the  readings  of  two  watt- 
meters which  are  connected  to  measure  the  total  power  by  the 
two-wattmeter  method. 

Refer  to  Fig.  99,  page  331.  From  equations  (40)  and  (41), 
page  331,  the  readings  of  the  two  wattmeters  are 

Pi  (wattmeter  in  line  1)  =  ILVL  cos  (0P°  -  30°) 
P2  (wattmeter  in  line  2)  =  ILVL  cos  (Op°  +  30°) 

*  Angles  of  lag  are  taken  positive  in  equations  (40)  (41). 


334  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Expanding  the  cosine  terms, 


Pi  =  lLVL    ^cos0p°  +  ±sin0p° 
I    2  z  } 


P2  = 
from  which 


~Y  cos  0P°  -   I  sin  6P° 


Pi  -  P2  =      sin  8P° 

Pi  +P2  ~  V3  cos  0/ 

tan0p°  =  Va^-qrl2  (42) 

fi  ~T  *2 

Since  the  tangent  of  the  angle  of  lag  between  the  phase  current 
and  phase  voltage  of  a  circuit  is  always  equal  to  the  ratio  of  the 
reactive  phase  power  to  the  true  phase  power,  it  is  obvious  from 
equation  (42)  that  \/3(Pi  —  PZ)  is  reactive  power.  For  bal- 
anced conditions,  the  square  root  of  three  times  the  difference 
of  the  readings  of  wattmeters  connected  to  measure  the  power 
of  a  three-phase  circuit  by  the  two-wattmeter  method  is  the 
reactive  power  of  the  circuit. 

An  Example  Involving  the  Use  of  the  Two-wattmeter  Method 
for  Measuring  the  Power  in  a  Balanced,  Three-phase,  Y-con- 
nected  Circuit  Containing  Third  Harmonics.  —  Each  branch  of  a 
Y-connected  load  consists  of  a  condenser,  an  air-core  induc- 
tance and  a  non-inductive  resistance,  in  series.  When  this 
circuit  is  connected  to  a  source  of  power  with  balanced  voltages 
which  contain  no  harmonics  of  higher  order  than  the  third,  each 
of  two  wattmeters,  connected  to  measure  power  by  the  two- 
wattmeter  method,  reads  15  kilowatts.  Each  voltage  between 
line  terminals  of  the  load  is  230  volts.  When  the  neutral  of  the 
Y-connected  load  is  connected  to  the  neutral  of  the  source  of 
power,  the  wattmeter  readings  and  the  voltages  between  the 
line  terminals  of  the  load  remain  unchanged,  but  there  is  a  current 
in  the  neutral  connection  of  60  amperes.  Under  this  condition, 
each  of  the  voltages  between  the  line  terminals  of  the  load  and 
the  neutral  point  is  150  volts.  What  are  the  resistance,  induct- 
ance and  capacitance  of  each  branch  of  the  load?  The  funda- 
mental frequency  of  the  impressed  voltage  is  60  cycles. 

Let  the  subscripts  1  and  3  attached  to  the  letters  P,  I  and  V 
indicate,  respectively,  fundamental  and  third-harmonic  power, 
current  and  voltage  per  phase. 


POWER  AND  POWER-FACTOR  335 

Since  the  line  voltages  of  a  balanced  three-phase  circuit  cannot 
contain  third  harmonics,  the  230  volts  impressed  across  the 
terminals  of  the  load  must  contain  the  fundamental  only. 
Therefore  the  fundamental  voltage  to  neutral  must  be 

oorj 

Vl  =  -    =  =  132.8  volts. 

V3 

Since  the  wattmeters  have  their  potential  coils  connected 
across  the  line  terminals  of  the  load,  the  voltages  impressed 
across  the  potential  coils  must  contain  the  fundamental  only. 
The  wattmeters  can  record  only  fundamental  power,  even  though 
their  current  coils  carry  third-harmonic  current,  since  their 
potential  coils  have  only  fundamental  voltage  impressed  across 
them.  There  can  be  no  power  developed  by  a  harmonic  in  the 
current  if  the  corresponding  harmonic  in  the  voltage  is  absent. 

Since  the  two  wattmeters  read  alike,  the  power-factor  of  the 
circuit  for  the  fundamental  alone  must  be  unity.     The  circuit 
must  therefore  be  in  resonance  for  the  fundamental. 
Pi  =  TVgn 

=  7,2 ^1 

1  n2  +  Z!2 

where  g\,  r\  and  x\  are  the  conductance,  resistance  and  reactance 
per  phase  for  the  fundamental.     For  resonance,  x\  is  zero  and 


_  Fi2  _          (132.8) 


Pi        15,000  +  15,000 

3 
=  1.76  ohms. 

The  third-harmonic  voltage  to  neutral  is 
y3  =  V050)2  -  (13278)' 

=  69.8  volts. 

Third-harmonic  current  per  phase  is  equal  to  one-third  of  the 
third-harmonic  current  in  the  neutral. 

73=-^=-;-  =20  amperes. 

o  o 

The  third-harmonic  impedance  per  phase  is 
23  =  -OFT  =  3.49  ohms. 

x,  =  V(3.49)2  -  (1.76)2 
=  3.01  ohms. 


336  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Let  XLI  and  xc\  represent  the  inductive  and  capacitive  reac- 
tances for  the  fundamental.  Let  XLS  and  xcz  represent  the  same 
quantities  for  the  third  harmonic.  Then 

XLI  —  Xci  =  0 

XLS  —  XC3  =  3xLi  —  ^Xci  =  3.01 

3Li(9  -  1)  =  9.03 
XLI  =  1.13  ohms. 
XGI  =  —1.13  ohms. 

L  =        -  =  =  0.00300  henry. 


106  106 

=  =  377XTI3  =  234?  microfarads' 


Measurement  of  the  Reactive  Power  of  a  Balanced  Three- 
phase  Circuit,  whose  Current  and  Voltage  Waves  are  Sinusoidal. 

If  the  potential  coils  of  two  wattmeters,  which  are  arranged  to 
measure  the  power  of  a  balanced  three-phase  circuit  whose 
current  and  voltage  waves  are  sinusoidal,  are  interchanged,  the 
reading  of  either  wattmeter  multiplied  by  the  square  root  of 
three  is  equal  to  the  reactive  power  in  the  circuit.  For  example, 
if  the  potential  coil  of  the  wattmeter  with  its  current  coil  in 
line  1,  Fig.  98,  page  326,  is  connected  between  lines  2  and  3 
instead  of  between  lines  1  and  3,  and  the  potential  coil  of  the 
wattmeter  with  its  current  coil  in  line  2  is  connected  between 
lines  1  and  3  instead  of  between  lines  2  and  3,  the  reading  of 
each  wattmeter  will  be 

ILVL  sin  Op 

If  the  potential  coils  of  the  two  wattmeters  are  reconnected 
as  stated  above,  it  will  be  seen,  by  referring  to  Fig.  99,  page 
331,  that  the  wattmeters  will  read  . 

Pi  =  /oJr32  cos  (90°  -  6P°) 

=  ILVL  sin  6P° 
Pi  =  /o2V,i  cos  (90°  +  0P°) 

=  ILVLsm  (-0P°) 

Reactive  power  =  37  phaseV  phase  sin  ephase 
=  SIpVp  sin  Bp 

L  sin  dp 


CHAPTER  XII 

UNBALANCED  THREE-PHASE  CIRCUITS 

Unbalanced  Circuits. — The  problems  involved  in  the  operation 
of  balanced  three-phase  circuits  or,  in  general,  of  balanced  poly- 
phase circuits  are  comparatively  simple  and  easy  of  solution. 
Except  when  there  is  interaction  between  phases,  such  problems 
may  be  treated  like  any  single-phase  problem  by  considering  a 
single  phase.  Although  in  practice  most  polyphase  circuits  are 
nearly  balanced,  many  cases  of  bad  unbalancing  exist.  Moreover, 
certain  types  of  polyphase  apparatus  are  particularly  sensitive 
to  any  unbalancing.  This  is  especially  true  of  the  rotary  con- 
verter, which  was  mentioned  on  page  304.  This  machine  will 
develop  bad  commutation  on  its  direct-current  side  and  certain 
other  bad  operating  features  when  operated  from  a  polyphase 
circuit  whose  voltages  are  even  moderately  out  of  balance.  For 
a  fixed  total  output,  the  copper  loss  in  all  types  of  polyphase 
apparatus  increases  with  the  amount  of  unbalancing.  For  these 
reasons  and  many  others,  any  method  which  will  simplify  the 
handling  of  problems  which  arise  when  dealing  with  unbalanced 
circuits  is  of  great  importance. 

A  method  of  great  power  for  handling  many  such  problems, 
which  although  not  new  has  come  into  prominence  during  the 
last  few  years  through  its  application  to  problems  arising  in 
connection  with  unbalanced  three-phase  circuits,  depends  upon 
the  fact  that  any  unbalanced  three-phase  system  of  sinusoidal 
vectors  may  be  resolved  into  three  component  systems  of  vectors.  * 
The  components  of  the  first  of  these  systems  are  identical.  They 
are  zero  when  the  vector  sum  of  the  three  original  vectors  is 
zero.  These  components  may  be  called  the  Uniphase  or  residual 
components.  The  components  of  the  second  system  together 

*  R.  E.  Oilman  and  C.  LeG.  Fortescue,  Trans.  Am.  Inst.  of  Elect.  Eng., 
vol.  XXXV,  1916,  p.  1329. 

C.  L.  Fortescue,  Trans.  Am.  Inst,  of  Elect.  Eng.,  vol.  XXXVII,  1918,  p. 
1027. 

W.  V.  Lyon,  Electrical  World,  June  5,  1920. 
22  337 


338  PRINCIPLES  OF  ALTERNATING  CURRENTS 

give  rise  to  a  balanced  three-phase  system  whose  phase  order  is 
the  same  as  the  phase  order  of  the  original  vectors.  The  com- 
ponents of  the  third  system  together  give  rise  to  a  balanced 
three-phase  system,  but  the  phase  order  of  this  system  is  oppo- 
site to  that  of  the  original  vectors.  In  other  words,  any 
unbalanced  three-phase  system  of  sinusoidal  voltages  may  be 
replaced  by  two  balanced  systems  of  sinusoidal  three-phase  volt- 
ages having  opposite  phase  orders  and  three  sinusoidal  uniphase 
voltage  vectors.  Any  unbalanced  system  of  three-phase  cur- 
rents, whose  wave  forms  are  sinusoidal,  may  be  similarly 
replaced. 

Direct,  Reverse  and  Uniphase  Components  of -Three-phase 
Vectors. — Let  Vi,  V%  and  F3  be  the  three  voltages  of  an  un- 
balanced three-phase  system  whose  yoltages  are  sinusoidal. 
Each  of  these  voltages  may  be  resolved  into  any  number  of 
sinusoidal  components  of  the  same  frequency,  but  since  there 
are  only  three  known  quantities,  namely,  Vi,  Vz  and  F3,  there 
can  be  only  three  independent  simultaneous  equations  to  deter- 
mine the  components.  Let  each  voltage  be  resolved  into  three 
components.  Then 

Vi  =  x    +  y    +  ~z  (1) 

F2  =  ax  +  by  +  cz  (2) 

F3  =  dx  +  ey  +  fz  (3) 

where  x,  y  and  z  are  the  three  components  of  Vi,  and  a,  b,  c,  d,  e 
and  /  are  complex  coefficients.  These  coefficients  must  be  so 
related  that  they  may  be  determined  from  equations  (1),  (2)  and 
(3),  which  are  the  only  possible  simultaneous  equations. 

Since  the  operation  of  motors  and  generators  is  well  understood 
under  balanced  conditions  and  can  be  easily  handled,  two  of  the 
coefficients,  b  and  e,  will  be  so  chosen  that  the  components  y,  by 
and  ey  form  a  balanced  three-phase  system  of  voltages,  whose 
phase  order  is  the  same  as  that  of  the  original  three  voltages. 
The  coefficients  c  and  /  will  be  so  fixed  that  the  components  z,  cz 
and  fz  also  form  a  balanced  three-phase  system  of  voltages,  but 
the  phase  order  of  this  system  of  voltages  will  be  opposite  to 
that  of  the  original  vectors.  The  remaining  coefficients,  viz., 
a  and  d,  will  be  fixed  so  that  the  components  x,  ax  arid  dx  are 
identical.  The  coefficients  a  and  d  are  therefore  each  unity. 


UNBALANCED  THREE-PHASE  CIRCUITS  339 

Then 

Fi  +  F2  +  F3  =  3x  +  (1  +  6  +  e)y  +  (1  +  c  +  /)«        (4) 
and 

y  +  5y  +  ~ey  =  0  (5) 

2  +  CZ  +  /*   =   0  (6) 

These  conditions  together  with  the  known  values  of  Fi,  F2 
and  V3  are  sufficient  to  definitely  fix  the  components.  Other 
systems  might  have  been  chosen,  but  this  would  complicate 
the  conditions  produced  by  unbalancing  rather  than  simplify 
them.  The  particular  advantage  of  the  systems  chosen  is  that 
two  are  balanced  systems,  for  which  the  conditions  are  well 
understood. 

When  the  conditions  are  such  that  the  vector  sum  of  the  three 
original  vectors  is  zero,  i.e.,  when 

Vi  +  v2  +  v3  =  o 

3x  must  also  be  zero,  since  y  +  by+ey  and  z  +  cz  +  Jz  are  each 
always  equal  to  zero. 

The  vector  sum  of  the  three-line  voltages  of  any  three-phase 
load  is  always  zero.  Hence  there  cannot  be  any  uniphase  com- 
ponents in  the  line  voltages  of  a  three-phase  load.  There  cannot 
be  any  uniphase  components  in  the  line  currents  of  either  a 
balanced  or  unbalanced  A-connected  load  or  in  the  line  or  phase 
currents  of  a  balanced  or  unbalanced  Y-connected  load  without 
a  neutral  connection,  since  the  vector  sum  of  the  currents  in 
each  case  must  be  zero.  There  probably  will  be  uniphase  com- 
ponents in  the  phase  voltages  of  an  unbalanced  Y-connected 
load  either  with  or  without  a  neutral  connection,  since  the  vector 
sum  of  these  voltages,  in  general,  is  not  zero.  It  is  possible, 
however,  to  have  an  unbalanced  Y-connected  load  in  which  the 
uniphase  components  are  zero.  Since,  in  general,  the  vector 
sum  of  the  phase  currents  in  an  unbalanced  A-connected  load 
is  not  zero,  these  currents  usually  will  contain  uniphase  com- 
ponents. In  this  case  also,  it  is  possible  to  have  a  load  in  which 
the  uniphase  components  are  zero.  There  will  be  uniphase 
components  in  the  line  and  phase  currents  of  an  unbalanced  Y- 
connected  load  with  a  neutral  connection  which  carries  current, 
since  the  vector  sum  of  the  phase  currents  and  therefore  the 
vector  sum  of  the  line  currents  is  not  zero. 


340  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Since  (1  +  b  +  e)y  and  (  1  +  c  +  /)z  are  each  equal  to  zero 
it  follows  that 

Vi  +  F2  +  ?3  =  3z 

i.Zt  +  F.+  V.  (?) 


That  is,  the  uniphase  component  in  each  phase  is  equal  to  one- 
third  of  the  vector  sum  of  the  three  original  vectors. 


Vi-x  =  y_  +   2  (8) 

V*-x  =  by  +  cz  (9) 

Vs  -  x  =  ey  +  fz  (10) 

The  vector  sum  of  the  drrecl_ajidj:everse  components  for  each 
phase  may  be  found  by  subtracting  the  uniphase  component 
vectorially  from  each  of  the  original  voltage  vectors. 

Instead  of  using  x,  y  and  z  as  the  components  of  Vi,  and  the 
coefficients  b,  c,  e  and  /,  it  will  be  simpler  and  more  convenient  in 
what  follows  to  use  the  subscripts  u,  d  and  r  with  the  letter  V 
to  indicate  the  uniphase,  the  direct-phase  and  the  reverse-phase 
components  respectively.  Using  this  notation 

yx  =  Vu+  7di  +  V_n  (11) 

F2    =    Vu   +    Vd2   +    Vr2  (12) 

F3    =    Vu   +    Fd3    +    Fr-3  (13) 

Vi  +  F2  +  V,    =  3Vu  (14) 

A  similar  relation  exists  among  the  currents  in  any  un- 
balanced three-phase  system.  For  example, 

/!=/«  +  l*i  +  In  (15) 

/2    =    In   +   Jrf2   +   Irt  (16) 

/3    =    In   +   Id»    +   Jrt  (17) 

/1    +   /2   +    /3    =    3/«  (18) 

Figure  101  shows  the  direct-  and  reverse-phase  components 
and  also  the  uniphase  components  of  the  currents  in  an  un- 
balanced Y-connected  three-phase  load  with  a  neutral  connection. 

The  direct-phase  components  are  shown  by  heavy  solid 
lines.  Heavy  dotted  lines  are  used  for  the  reverse-phase  com- 
ponents. The  uniphase  components  are  shown  by  solid  lines. 
Dot-and-dash  lines  are  used  for  the  original  vectors. 


UNBALANCED  THREE-PHASE  CIRCUITS 


341 


The  three  components,  i.e.,  the  direct-phase,  reverse-phase 
and  uniphase  components,  of  each  of  the  original  vectors  of  an 
unbalanced  three-phase  system  of  voltages  or  currents  may  be 
determined  either  analytically  or  graphically  from  the  known 
magnitudes  and  phase  relations  of  the  original  three-phase  vectors. 


FIG.  101. 


Determination  of  the  Direct,  Reverse  and  Uniphase  Compo- 
nents of  a  Three-phase  System.  —  Consider  the  three  voltage 
vectors  given  by  equations  (11),  (12)  and  (13).  It  has  already 
been  shown  that  the  uniphase  components  are 


Vu 


T3 


(19) 


Also 


V*i  +  Vdz  +  Vu  =  0  (20) 

Vrl   +    VrZ   +    V*    =    0  (21) 

Let  V's  with  primes  indicate  the  vectors  obtained  by  subtract- 
ing the  uniphase  components  from  each  of  the  given  three-phase 
vectors.  Then  if  the  phase  order  of  the  direct-phase  components 
is  clockwise 

"Vz*  —  Vz  —  Vu  = 
TV  =  F3  -  Vu  = 


+  V,,  =  YJI  -240°  +  Fr,  +240°  (24) 


342 


PRINCIPLES  OF  ALTERNATING   CURRENT'S 


If  the  system  is  balanced,  the  uniphase  components,  Vv,  and 
each  of  the  reverse-phase  components  are  zero. 

Let  each  of  the  vectors  in  equation  (23)  be  rotated  in  a  clock- 
wise direction  through  120  degrees  by  applying  the  operator 
1-120°.  Then 

TV  ~  120°  =  Vdi  -240°   +  Vn  J5!  (25) 


Subtracting  equation  (25)  from  equation  (22)  gives 


TV  _ 


-120° 


-  Fj-240 


-240 


(26) 

The  voltages   {  TV  -  TV  "120°  |,  FjO°  and  Vdl  -240°  are 
shown  in  Fig.  102. 


FIG.   102. 

By  referring  to  Fig.  102  it  will  be  seen  that 
TV  -  TV |- 120°  1  =  {TV  +  TVI+6Q°1  =  A/3Vdil-3Q°(27) 
Hence 

(28) 
(29) 


Since  Fi'  =Fdi+Fr,  (Equation  (22)) 
Vn  =  Fi'  -  Vdi 


(30) 


to 


UNBALANCED  THREE-PHASE  CIRCUITS  343 

Referring  to  Fig.  103  it  will  be  seen  that  equation  (30)  reduces 

1 


--tvi'l-ao"  +  TV -9Q° 


Having  determined  the  direct-  and  reverse-phase  components, 
Vdi  and  Fri,  for  phase  1,  the  direct-  and  reverse-phase  systems 
of  components  may  be  found  by  applying  the  proper  operators  to 

Vdi  and  Vn. 

Vdi  =  Fdl|-0°  (33) 

Vdz  =  7^1-120°  (34) 

V*  =  V*  1-240°  (35) 

Vn  =  Vn  1+0°  .(36) 


yr2  =  Vn  1  +  120 


7rll+240 


(37) 
(38) 


344 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


From  equation  (28)  it  is  obvious  that  to  find  the  magnitude 
of  the  direct-phase  component,  Vdi,  for  phase  1,  F2'  must  be 
rotated  in  a  positive  or  counter-clockwise  direction  through  60 
degrees,  added  to  Vi  and  the  result  divided  by  the  square  root 
of  three.  The  correct  phase  position  of  this  direct-phase  com- 
ponent is  found  by  rotating  it,  as  just  found,  through  30  degrees 
in  a  positive  or  counter-clock-wise  direction. 


FIG.  104. 


FIG.   105. 


From  equation  (31)  it  is  obviousjhat  to  find  the  magnitude 
of  the  reverse-phase  component,  Vri,  for  phase  1,  F2'  must 
be  rotated  in  a  negative  or  clockwise  direction  through  60 
degrees,  added  to  V\  and  the  result  divided  by  the  square  root 
of  three.  The  correct  phase  position  of  this  reverse-phase 
component  is  found  by  rotating  it,  as  just  found,  through  30 
degrees  in  a  negative  or  clockwise  direction. 

The  method  of  finding  the  direct-  and  reverse-phase  compo- 
nents for  phase  1  is  illustrated  in  Figs.  104  and  105.  A  graphical 
determination  of  the  direct-  and  reverse-phase  components  is 
sufficiently  accurate  for  many  purposes.  When  the  vector 


UNBALANCED  THREE-PHASE  CIRCUITS 


345 


expressions  for  the  three  vectors  which  are  to  be  resolved  into 
direct-  and  reverse-phase  components  are  known,  an  analytical 
solution  is  simple  and  will  require  little  if  any  more  time  than 
the  graphical  solution. 

A  Simple  Graphical  Construction  for  Finding  the  Direct-  and 
Reverse -phase  Components  of  a  Three-phase  Circuit  whose 
Vectors  are  Sinusoidal  and  Contain  no  Uniphase  Components.— 
When  there  are  no  uniphase  components  in  the  currents  or 
voltages  of  a  three-phase  circuit,  whose  currents  and  voltages 
are  sinusoidal,  and  only  the  magnitudes  of  the  currents  or  voltages 


FIG.  107. 


are  known,  the  direct-  and  reverse-phase  components  may  be 
found  by  a  very  simple  graphical  construction.  *  The  vector  sum 
of  the  line  voltages  of  a  three-phase  circuit  is  zero.  They  there- 
fore can  contain  no  uniphase  components.  Also  the  vector  sum 
of  the  line  currents  of  a  three-phase  A-connected  circuit,  or  Y-con- 
nected  circuit  without  a  neutral,  must  be  zero.  They  can  contain 
no  uniphase  components. 

Suppose  the  magnitudes  of  the  line  voltages  of  a  three-phase 

*See  Unbalanced  Three-phase  Circuits,  W.  V.  Lyon,  Electrical  World, 
June  5,  1920. 


346  PRINCIPLES  OF  ALTERNATING  CURRENTS 

circuit,  whose  voltages  are  sinusoidal,  are  known.  Since  their 
vector  sum  must  be  zero,  they  must  form  the  sides  of  a  closed 
triangle,  as  shown  in  Fig.  106,  where  Viz',  VM',  and  Vsi  are  the 
voltages.  Construct  an  equilateral  triangle  2-4-3  on  Vzs'  as  a 
base.  The  side  2-4  of  this  triangle  is  Vw  rotated  through  60 
degrees  in  a  positive  direction.  From  equation  (28),  page  342 
the  diagonal  1-4  =  V\<!  divided  by  the  square  root  of  three  is 
equal  to  the  magnitude  of  the  direct-phase  component  of  phase 
1-2. 

Draw  the  isosceles  triangle  1-5-4  on  1-4  as  a  base,  with  30- 
degree  angles  at  1  and  4,  as  shown  in  Fig.  106.     Then 


»  s 

=  Vd-n  =  direct-phase  component  for 
phase  1-2. 

A  similar  construction,  shown  in  Fig.  107,  gives  the  reverse- 
phase  component,  W-  12,  of  phase  1-2. 

Example  of  the  Calculation  of  the  Direct,  Reverse  and  Uniphase 
Components  of  the  Currents  in  an  Unbalanced,  Y-con- 
nected,  Three-phase  Circuit  whose  Currents  are  Sinusoidal.  — 
Measurements  show  that  the  line  currents  in  an  unbalanced,  Y- 
connected,  three-phase  circuit,  with  neutral  connection,  whose 
currents  are  sinusoidal,  are 

701  =  100|  ~  0°  amperes. 

702  =  1001-100°  amperes. 

703  =    75|^250^  amperes. 

What  are  the  direct,  reverse  and  Uniphase  components  of  the 
currents? 

7oi  =  7]  =  100!-_0°_      =  lOOjcos  (-0°)  +  j  sin  (-0°)} 

=  100  +  JO 
1  02  =  1  2  =  100  -100°    =  1  OOf  cos  (-100°)  +j  sin  (-100°)} 

=  -17.36  -  J98.48 
7o3  =  73  =   75    -250°    =  75jcos  (-250°)  +  j  sin  (-250°)} 

=  -25.65+  J70.48 


UNBALANCED  THREE-PHASE  CIRCUITS  347 


j  n   _ 

1 


~In   _  7i    +  72    +  1 


=  (100  -  17.36  -  25.65)  +  j(0  -  98.48  +  70.48) 

3 
=  19.00  -  J9.33 

7i'  =  Idi  +  In  =  7i  -7«    =  (100  -  JO)  -  (19.00  -  j'9.33) 

=  81.00  +J9.33 

It      =    Idt   +   7r2    =    1  2    ~    In     =    ("17.36    -  J98.48) 

-  (19.00  -  J9.33) 
=  -36.36  -j'89.15 

/3;  =  la  +  7rt  =  h-L    =  (-25.65  +  J70.48) 

-  (19.00  -  J9.33) 
=  -44.65  +./79.81 

It  is  obvious  from  equation  (29),  page  342,  that 

1 
Idi  =  —;=  j/i'(cos  30°  +  j  sin  30°)  +  //(cos  90°  +  j  sin  90°)  } 

=  ^{(81.00  +  J9.33X0.866  +  j  0.500) 

+  (-36.36  -J89.15)  (0.000  +  jl.000)| 
=  -±={  (65.48  +  J48.58)  +  (89.15  -  J36.36)} 

=  89.3  +J7.05 
From  equation  (32),  page  343, 

Jri  =  71  1  1  (cos  30°  -  /sin  30°)  +  72'(cos  90°  -  j  sin  90°)  } 


=  -|{  (81.00  +J9.33)  (0.866  -  J0.500)  } 

H-  (-36.36  -  j89.  15)  (0.000  -  jl.000)j 
=  ^{(74.81  -  J32.42>+  (-89.15  +J36.36)| 

=  -8.28  +J2.27 
Then 

7ui  =  19.00  -  J9.33 
7M2  =  19.00  -  J9.33 
7U3  =  19.00  -  J9.33 


348  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Idl  =  89.3    +  J7.05 

Id2  =  (89.3  +  J7.05)(cos  120°  -  j  sin  120°) 

=  -38.55  -  J80.86 
7d8  =  (89.3  +  J7.05)  (cos  240°  -  j  sin  240°) 

=  -50.76  +  J73.81 
7ri  =  -8.28  +  J2.27 
7r2  =  (-8.28  +j2.27)(cos  120°  +  j  sin  120°) 

=  2.174  -  J8.305 
7r3  =  (-8.28  +  J2.27)(cos  240°  +  j  sin  240°) 

=  6.106  +  J6.035. 

Mutual-induction  between  a  Three-phase  Transmission  Line 
and  a  Neighboring  Telephone  Line. — The  vector  sum  of  the 
direct-phase  components  and  the  vector  sum  of  the  reverse- 
phase  components  of  the  currents  of  a  transmission  line  are  zero. 
Therefore,  if  the  conductors  of  a  three-phase  transmission  line 
could  be  equidistant  from  the  conductors  of  a  telephone  line, 
which  runs  parallel  to  the  transmission  line,  these  component 
currents  could  produce  no  inductive  effects  on  the  telephone  line. 
Although  it  is  impossible  to  have  the  conductors  of  a  transmis- 
sion line  equidistant  from  the  conductors  of  a  telephone  line, 
they  can,  on  the  average,  be  made  equidistant  by  properly  trans- 
posing them.  By  proper  transposition,  the  inductive  effects  of 
the  direct-  and  reverse-phase  components  may  be  made  zero. 
This  is  not  true  of  the  uniphase  components.  These  are  all  in 
phase  and  no  amount  of  transposition  of  the  conductors  of  the 
transmission  line  will  alter  their  inductive  effects  on  the  telephone 
line.  To  get  rid  of  the  inductive  effects  of  the  uniphase  com- 
ponents, the  telephone  line  must  be  transposed.  The  uniphase 
components  or  residuals,  as  they  are  called  by  telephone  en- 
gineers, play  a  very  important  part  in  the  interference  effects 
produced  on  telephone  lines  by  unbalanced  transmission  lines 
which  are  operated  with  grounded  neutrals. 

The  whole  analysis  of  the  interference  between  transmission 
lines  and  neighboring  telephone  lines  is  much  simplified  by  resolv- 
ing the  currents  in  the  transmission  line  into  direct-phase,  reverse- 
phase  and  uniphase  components. 


UNBALANCED  THREE-PHASE  CIRCUITS  349 

Power  in  an  Unbalanced  Three-phase  Circuit  when  the  Cur- 
rents and  Voltages  are  Sinusoidal.  —  The  power  in  a  three-phase 
circuit  is 

Po  =  Pi  +  P2  +  Pa  (39) 

where  PI,  P2  and  P3  are  the  powers  in  phases  1,  2  and  3  respec- 

tively, 

P!  =  Fi/i  cos  0i  (40) 

P2  =  F2/2  cos  02  (41) 

PS  =  F3/3  cos  63 

where  the  F's,  I's  and  0's  represent  phase  values  of  voltage, 
current  and  power-factor  angle. 

If  Vi  and  1  1  are  each  resolved  into  direct-phase,  reverse-phase 
and  Uniphase  components,  the  expression  for  PI  becomes 

Pi    =   Vdld  COS  07,;  +   Vrlr  COS   B^r\  +   VJU  COS  O 
+   Vdlr  COS  07?;  +   Fd/tt  COS  0 

+  vrid  cos  0/;;  +  vriu  cos  07: 

+  Fu/d  cos  07ud;  +  VJr  cos  0 

where  Fd,  /<*,  Vr,  Ir,  Vu  and  7U  represent  the  numerical  values 
of  the  direct-phase,  reverse-phase  and  Uniphase  components  of 
the  voltages  and  currents.  The  limits  on  the  phase  angles,  0, 
indicate  to  which  phase  they  refer. 

Expressions  which  are  similar  to  equation  (42)  may  be  written 
for  P2  and  P3. 

When  the  expressions  for  PI,  P2  and  P3  are  added  to  give  the 
total  power,  P0,  it  will  be  found  that  the  sums  of  the  terms  involv- 
ing unlike  components  are  zero.  For  example:  the  sum  of  the 
three  following  terms  is  zero. 

Vdlr  COS    0^;   +    Vdlr  COS  07?!  +    VJr  COS  0^    = 

Fd/r(cos  07:;  +  cos  07:;  +  cos  07::)  =  o  (43) 

The  components  in  equation  (43)  are  plotted  in  Fig.  108. 
By  referring  to  Fig.  108,  it  will  be  seen  that  equation  (43) 
may  be  written  in  the  following  form, 
TVr{cos  a  +  cos  (120°  +  a)  +  cos  (120°  -  «)}  = 
Fd/r(cos  a  +  cos  120°  cos  a  —  sin  120°  sin  a 

+  cos  120°  cos  a  +  sin  120°  sin  a]  = 


(42) 


T7   r   J  1  A/3    .  1  ,    A/3    - 

v  */-•{  cos  a  —  -cos  a—  — —  sma  —  -cosa  +  -^—  sin  eel    =  0 
Z  2  2  Z 


350  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  sum  of  the  other  terms  involving  components  which  are 
unlike  may  similarly  be  shown  to  be  zero.  Therefore 

Po  =  3V dld  cos  Ojl  +  3VTIr  cos  0£  +  3VUIU  cos  evT:      (44) 

The  total  power  in  an  unbalanced  three-phase  circuit,  which 
has  no  uniphase  components,  is  equal  to  the  sum  of  the  powers 
developed  by  the  direct-  and  reverse-phase  components. 
When  reverse-phase  components  are  present  in  the  currents 

and  voltages  of  a  three-phase  circuit, 
i.e.,  when  the  circuit  is  unbalanced, 
the  power  developed  in  at  least  one 
phase  is  greater,  and  in  at  least  one 
other  phase  is  less  than  the  average 
power  per  phase.  If  there  were  no 
reverse-phase  components,  the  power 
developed  by  all  phases  would  be  the 
same.  The  effect  of  the  reverse- 
phase  components  is  to  transfer 

10o. 

power  from  one  phase  to  another. 
The  principle  of  the  phase  balancer  depends  on  this  trans- 
fer of  power  from  one  phase  to  another  by  reverse-phase 
components. 

Phase  Balancer. — It  is  necessary,  in  certain  cases,  for  central 
stations  to  supply  large  amounts  of  single-phase  power  for  special 
purposes,  such  as  for  the  operation  of  electric  railways  using 
single-phase,  alternating-current,  series  motors.  This  single- 
phase  load  not  only  badly  unbalances  the  voltage  of  the  system 
supplying  it,  but  also  very  much  decreases  the  permissible  output 
from  the  generating  equipment.  The  operation  of  certain  types 
of  polyphase  apparatus,  notably  the  rotary  converter,  is  difficult 
on  circuits  whose  voltage  is  out  of  balance. 

The  impedance  of  a  polyphase  rotating  machine,  such  as  a 
synchronous  motor  provided  with  a  damping  winding  or  an 
induction  motor,  is  much  less  for  the  reverse-phase  components 
of  an  unbalanced  system  of  voltages  than  for  the  direct-phase 
components.  Such  a  machine,  when  connected  to  a  circuit 
having  unbalanced  voltages,  will  take  reverse-phase  currents 
which  are  large  compared  with  the  direct-phase  currents.  The 
reverse-phase  currents  will  be  substantially  opposite  in  phase  to 


UNBALANCED  THREE-PHASE  CIRCUITS  351 

the  reverse-phase  currents  in  the  load  which  causes  the  unbal- 
anced voltages.  For  this  reason,  a  synchronous  motor  with 
damping  winding  or  an  induction  motor  will  partially  restore 
the  condition  of  balance  of  currents  in  an'  unbalanced  circuit 
to  which  it  is  connected,  but  such  a  machine  alone  cannot 
restore  the  condition  of  complete  balance  of  currents. 

A  type  of  apparatus  has  been  developed  within  the  last  few 
years  by  means  of  which  complete  balance  may  be  restored  to 
an  unbalanced  three-phase  circuit.  This  is  accomplished  by 
taking  from  the  unbalanced  circuit  reverse-phase  currents  equal 


G=  Source  of  Power 

S=  Synchronous  Motor  Coupled  to  B 

B==  Pha=e  Balancing  Alternator 

FIG.  109. 

in  magnitude  but  opposite  in  phase  to  those  caused  by  the  exist- 
ing unbalanced  load.  Since  unbalanced  voltages  are  usually  due 
to  unbalanced  line  drops  caused  by  an  unbalanced  load,  balancing 
the  currents  of  a  circuit  will  usually  balance  the  voltages. 

The  Phase  Balancer,  as  the  machine  is  called,  consists  of  a 
synchronous  motor,  with  a  low  impedance  damper,  driving  a 
three-phase  synchronous  alternator.  One  phase  of  the  alter- 
nator is  connected  in  series  with  each  phase  of  the  synchronous 
motor,  but  the  phase  orders  of  the  alternator  and  motor  are 
opposite.  The  alternator  impresses  reverse-phase  voltages  of 
such  magnitude  and  phase  that  the  motor  takes  reverse-phase 
currents  from  the  line  which  are  equal  and  opposite  to  those 
in  the  unbalanced  load.  An  automatic  device  is  used  to  adjust 
the  magnitude  and  phase  of  reverse-phase  voltages  of  the  phase- 
balancing  alternator.  The  magnitude  of  the  voltages  is  ad- 
justed by  means  of  the  field  excitation  of  the  alternator.  Their 
phase  is  varied  by  rotating  the  magnetic  axis  of  the  field  with 
respect  to  the  field  structure  by  the  use  of  two  independently 
excited  field  windings  displaced  ninety  degrees  from  each  other 
on  the  field  structure. 


352 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


The  development  of  the  phase-balancer  would  have  been 
improbable  without  the  knowledge  that  an  unbalanced  three- 
phase  load,  with  sinusoidal  currents  and  voltages  and  without 
a  neutral  connection,  could  be  resolved  into  two  balanced  three- 
phase  loads  having  opposite  phase  orders. 

The  schematic  diagram  of  connections  for  a  phase  balancer 
to  balance  currents  is  shown  in  Fig.  109. 

Copper  Loss  in  an  Unbalanced  Three-phase  Circuit  in  Terms 
of  the  Direct,  Reverse  and  Uniphase  Components  of  the  Currents. 
The  copper  loss  in  phase  1  is 

-Pi  copper    =    (7i  phase)2    X   T\ 

The  direct,  reverse  and  uniphase  components  of  the  currents 
are  shown  in  Fig.  110. 


If  Ir\  and  Iu  are  each  resolved  into  two  quadrature  components 
with  respect  to  hi,  it  will  be  seen,  by  referring  to  Fig.  110,  that 

Pi  copper   =    I  [Idl  +  Irl  COS  «  +  Iu  COS  ft]2 

+  [Iri  sin  a  +  Iu  sin  ft}2}rl 
=    [Idi2  +  In2  +  7w2]n  +  2[/di/ri  cos  a  +  Idju  cos  ft 

+  IrJu  cos  a  cos  ft  +  Irllu  sin  a  sin  ft]ri         (45) 
Similarly,  by  resolving  Ir2  and  Iu  each  into  two  quadrature 
components  with  respect  to  7<f2  gives 

I    2  copper    =    (72  phase)      X    7*2 

=  {[/d2  +  Ir2  cos  (240°  -  «)  +  /u  cos  (120°  -  ft)]2 

+[7r2  sin  (240°  -  a)  +  7M  sin  (120°  -  /3)]2}r2 
=  [7d22  +  7r22  +  7M2]r2  +  2[7d27r2  cos  (240°  -  a) 
+  7d27M  cos  (120°  -  ft) 

+  7r27M  cos  (240°  -  a)  cos  (120°  -  ft) 

4-  7r27u  sin  (240°  -  a)  sin  (120°  -  0)]r2       (46) 


UNBALANCED  THREE-PHASE  CIRCUITS  353 

Resolving  7r3  and  7M  each  into  quadrature  components  with 
respect  to  Ids  gives 

•*   3    copper    =    (.-'3  copper)      X   ?*3 

=  {[7d3  +  /rs  cos  (120°  -«)+/,  cos  (240°  -  0)]2 

+  [7r3  sin  (120°  -  a)  +  7U  sin  (240°  -  0)]2}r3 
=  [7d32  +  7r32  +  7w2]r3  +  2[7d37r3  cos  (120°  -  a) 
+  IdJu  cos  (240°  -  0) 

+  7r37u  cos  (120°  -  a)  cos  (240°  -  0) 

+  7r37u  sin  (120°   -  a)  sin  (240°   -  0)]r,      (47) 

Since  I*i  =  7«*2  =  Ids  and  Ir\  =  7r2  =  7r3  in  magnitude,  the  sub- 
scripts 1,  2  and  3  may  be  omitted,  since  the  currents  and  voltages 
enter  only  in  magnitude  in  equations  (45),  (46)  and  (47). 

When  the  resistances  r*i,  r2  and  r3  are  equal,  as  would  be  the 
case  for  a  motor  or  a  generator,  the  equation  for  the  total  power 
due  to  copper  loss  becomes 

Po  =  Pi  +  P2  +  P3 

=  3{7d2  +  /r2  +  7M2}r 

+  2{7d7r[cos  a  +  cos  (120°  -  a)  +  cos  (240°  -  a)] 
+  7d7w[cos  0  +  cos  (120°  -  0)  +  cos  (240°  -  0)] 

+  7r7u[cos  a  cos  0  +  cos  (240°  -  a)  cos  (120°  -  0) 

+  cos  (120°  -  a)  cos  (240°  -  0)] 
+  7r7M[sin  a  sin  0  +  sin  (240°  -  a)  sin  (120°  -  0) 
+  sin  (120  -  a)  sin  (240°  -  0)]}r  (48) 

The  sum  of  the  cosines  of  any  three  angles  which  differ  by  120 
degrees  is  zero.  The  second  and  third  terms  of  equation  (48) 
are  therefore  zero.  The  last  two  terms  are  also  zero  for  the 
same  reason.  These  terms  may  be  written  in  the  following  form 

7r7u{cos  [(a  -  0)]  +  cos  [(a  -  0)   -  120°] 

+  cos  [(a  -  0)  -  240°]   =  0 

The  total  copper  loss,  when  the  resistances  of  all  three  phases 
are  equal,  as  they  would  be  in  the  case  of  motors  or  generators,  is 
equal  to 

Po  =  3(7d2  +  7r2  +  7u2)r  (49) 

That  is,  the  total  copper  loss  in  an  unbalanced  three-phase 
system  whose  phase  resistances  are  equal,  is  equal  to  the  sum  of 
the  squares  of  the  direct-phase,  reverse-phase  and  uniphase  com- 
ponents, of  the  currents  multiplied  by  the  phase  resistance. 

23 


354  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Effect  of  Impressing  an  Unbalanced  Voltage  on  a  Three-phase 
Alternating-current  Motor  or  Generator. — The  resolution  of  the 
voltages  and  currents  of  three-phase  motors  and  generators, 
which  operate  under  unbalanced  conditions,  into  direct-phase  and 
reverse-phase  components  (uniphase  components  do  not  ordinarily 
exist  in  motors  or  generators)  is  one  of  the  most  powerful  methods 
of  attacking  the  problems  involved  in  determining  the  effect 
of  the  unbalanced  conditions  on  operation.  Equation  (49)  shows 
that  the  direct-  and  reverse-phase  components  of  the  currents, 
and  also  the  uniphase  components  if  they  exist,  contribute  to 
the  total  copper  loss.  The  torque  developed  in  a  motor  or 
generator  by  the  reverse-phase  components  of  the  armature 
currents  is  opposite  to  that  developed  by  the  direct-phase  com- 
ponents and  therefore  subtracts  from  the  total  net  output.  The 
presence  of  reverse-phase  components  in  the  armature  currents 
not  only  increases  the  total  armature  copper  loss  in  a  machine 
but  it  also  subtracts  from  the  total  power  developed.  The  torque 
developed  per  ampere  of  the  reverse-phase  components  of  the 
armature  currents  is  much  smaller  than  the  torque  developed 
per  ampere  of  the  direct-phase  components  of  those  currents. 

If  the  direct-  and  reverse-phase  components  of  the  voltage 
impressed  on  a  circuit  are  known,  the  direct  and  reverse-phase 
component  currents  may  be  calculated,  provided  the  resistances, 
inductances  and  capacitances  of  the  circuit  are  constant,  or  their 
magnitudes  are  known  for  both  components  of  the  currents.  In 
a  motor  or  generator,  the  apparent  impedance  for  the  reverse- 
phase  components  of  the  currents  is  quite  different  from  the 
apparent  impedance  for  the  direct-phase  components.  Equal 
reverse-  and  direct-phase  components  in  the  impressed  voltages 
will  produce  reverse-  and  direct-phase  components  in  the  currents 
of  quite  different  magnitudes. 

The  study  of  the  operation  of  the  three-phase  induction  motor 
under  unbalanced  conditions  of  impressed  voltage,  or  even  when 
it  is  operated  single  phase,  may  be  attacked  with  great  advantage 
from  the  standpoint  of  direct-  and  reverse-phase  components. 
The  study  of  the  single-phase  induction  motor  may  be  simplified 
by  considering  the  single-phase  current  to  be  resolved  into 
direct-  and  reverse-phase  components  of  a  three-phase  system. 


CHAPTER  XIII 
REACTANCE  OF  A  TRANSMISSION  LINE 

Reactance  of  a  Single-phase  Transmission  Line. — When  all  of 
the  flux  of  a  circuit  does  not  link  all  of  the  current,  a  more 
fundamental  conception  of  flux  linkages  than  flux  linkages  with 
turns  or  with  conductors  must  be  used  in  determining  the 
self-  or  mutual-inductance  of  the  circuit.  Such  a  case  arises 
when  the  portion  of  the  self-inductance  of  a  conductor  which  is 
due  to  the  flux  within  the  conductor  is  determined.  A  similar 
case  occurs  in  the  determination  of  the  portion  of  the  mutual- 
inductance  of  a  circuit  which  is  due  to  the  flux  which  lies  within 
the  conductors.  In  such  cases  it  is  necessary  to  consider  the 
flux  linkages  with  respect  to  current.  When  all  of  the  flux  con- 
sidered lies  without  the  conductors,  or  when  that  portion  of 
the  flux  within  the  conductors  may  be  neglected,  it  makes  no 
difference  in  the  final  result  whether  the  linkages  are  taken  with 
respect  to  the  current  or  merely  with  respect  to  the  circuit. 
Since  flux  linkages  with  current  must  be  used  in  a  portion  of 
what  follows,  they  will  be  used  throughout  for  uniformity.  As 
all  currents  and  fluxes  are  vectors  they  must  be  considered 
either  vec tonally  or  as  instantaneous  values.  The  sense  in 
which  they  are  considered  in  what  follows  is  clearly  indicated 
by  the  notation  which  has  been  used  throughout  this  book. 
Small  letters  indicate  instantaneous  values.  Capital  letters 
with  a  dash  over  them  indicate  vectors. 

Consider  a  single-phase  transmission  line  consisting  of  two 
straight,  parallel,  cylindrical  conductors,  A  and  5,  of  circular 
cross  section,  each  of  r  centimeters  radius.  Let  the  distance 
between  the  axes  of  the  conductors  be  D  centimeters. 

The  intensity,  3C,  of  the  magnetic  field  due  to  a  long,  straight, 
cylindrical  conductor  of  circular  cross  section  at  a  point  p 
outside  the  conductor  and  at  a  perpendicular  distance  x  from  its 
axis  is 

5C  =  —  gausses  (1) 

x 

355 


356  PRINCIPLES  OF  ALTERNATING  CURRENTS 

where  i  is  the  current  in  the  conductor  in  abamperes.  Equation 
(1)  assumes  that  the  current  distribution  over  the  cross  section 
of  the  conductor  is  uninfluenced  by  any  magnetic  field  other 
than  that  produced  by  the  current  in  the  conductor  itself.  The 
effect  is  the  same  as  if  the  current  were  all  concentrated  in 
the  axis  of  the  conductor. 

For  a  transmission  line  carrying  current  of  usual  frequency 
and  having  the  usual  spacing  between  conductors  as  compared 
with  their  diameter,  the  effect  of  the  current  in  any  conductor 
on  the  current  distribution  in  the  other  conductors  may  be 
neglected.  This  effect  cannot  be  neglected  in  cables,  since  the 
separation  between  the  conductors  of  cables  is  small  compared 
with  the  diameter  of  the  conductors. 

The  flux  density  at  the  point  p  is 

(B  =  3Cju  =  —/A  gausses  (2) 

3& 

=  —  gausses  (3) 

x 

since  the  permeability  /*  is  unity. 

Consider  the  single-phase  line  shown  in  Fig.  111. 

The  total  inductance  of  the 
conductor  A  will  consist  of  three 
parts,  namely: 

(a)  That  due  to  the  flux  link- 
ages resulting  from  the  flux 
produced  within  the  conductor 
by  its  own  current,  i.e.,  by  the 
current  iA- 

Flo   m  (6)  That  due  to  the  flux  link- 

ages resulting  from  that  portion 

of  the  flux,  produced  by  the  current  iA,  which  lies  between  the 
surface  of  the  conductor  and  infinity,  i.e.,  between  the  limits  r 
and  infinity. 

The  parts  contributed  by  (a)  and  (6)  together,  per  unit  current, 
constitute  the  self -inductance  of  the  conductor  A . 

(c)  That  due  to  the  flux  linkages  resulting  from  so  much  of  the 
flux  produced  by  the  current  iB  in  conductor  B  as  links  conductor  A . 
This,  per  unit  current,  is  the  mutual  inductance  of  B  on  A . 


REACTANCE  OF  A  TRANSMISSION  LINE  357 

PART  a. — Due  to  skin  effect,  the  current  density  within  the 
conductors  will  not  be  uniform,  but  for  ordinary  frequencies  and 
sizes  of  conductor  used  in  power  transmission,  no  great  error  will 
be  produced  in  the  total  inductance  per  conductor  by  assuming 
uniform  current  density.  In  fact,  the  part  of  the  inductance 
due  to  the  flux  linkages  within  the  conductor  may  be  neglected 
without  introducing  any  appreciable  error  for  high-voltage 
transmission  lines  with  the  usual  separation  between  conductors. 
Low  voltage  lines  are  usually  so  short  that  the  line  inductance 
ceases  to  be  an  important  factor.  With  a  ratio  of  D  to  r  as  low 
as  50,  the  flux  linkages  caused  by  the  flux  within  the  conductors 
account  for  only  about  seven  per  cent  of  the  total  line  inductance. 

Assuming  uniform  current  density  within  the  conductors,  the 
current  density  at  any  point  within  the  conductor  A  is 

p  =  —  abamperes  per  square  centimeter.  (4) 

TTT 

The  current  in  that  portion  of  the  conductor  whose  radius  is 
x  centimeters  is 

7TX2.  X2.         ,  /cx 

— ^IA  =  -jiA  abamperes.  (5) 

The  field  intensity  at  a  distance  x  centimeters  from  the  axis 
of  the  conductor  due  to  this  portion  of  the  current  is 

2x*.         2x.  ft~ 

^IA  ==  -^IA  gausses. 

The  field  intensity  within  a  hollow  cylindrical  conductor  due 
to  the  current  it  carries  is  zero.  Hence  the  field  at  a  distance  x 
from  the  axis  of  a  conductor  of  circular  cross  section,  due  to  the 
current  in  that  portion  of  the  conductor  without  x,  is  zero.  This 
assumes  that  the  current  is  either  uniformly  distributed  or  has 
the  same  density  at  equal  distances  from  the  axis  of  the  conductor. 

The  flux  through  an  element  of  the  conductor  of  radius  x  centi- 
meters, width  dx  centimeters  and  length  one  centimeter  is 

2x 

—jiAu'dx  maxwells,  (7) 

where  p  is  the  permeability  of  the  conductor. 

This  flux  does  not  link  all  of  the  current  in  the  conductor,  but 

x2 
only  that  portion  -jiA  which  is  within  the  radius  x.     The  total 


358  PRINCIPLES  OF  ALTERNATING  CURRENTS 

flux  linkages   with   the   current   per   centimeter   length   of   the 
conductor  are    . 

For  non-magnetic  conductors,  such  as  are  used  for  transmission 
lines,  this  reduces  to 


PART  6. — The  flux  density  due  to  the  current  iA  at  any  point 
p  outside  the  conductor  A  and  distant  x  from  the  axis  of  the 
conductor  is  (see  Fig.  111). 

(B  =  -^gausses.  (10) 

x 

The  flux  through  an  annulus  of  radius  x  centimeters,  width 
dx  centimeters  and  length  one  centimeter  measured  parallel  to 
the  axis  of  the  conductor  is 

($>dx  =  -~  dx  maxwells.  (11) 

This  links  the  current  iA  in  the  conductor  A.  The  total  flux 
linkages  with  the  current  iA  in  conductor  A  are  therefore 

2iA2  oc 

— —  dx  =  2tA2loge-  (12) 


per  centimeter  length  of  conductor  A. 

PART  c.  —  The  flux  due  to  the  current  ia,  in  conductor  B,  which 
links  the  current  iAj  in  conductor  A,  includes  all  the  flux  pro- 
duced by  the  current  iB  which  lies  between  a  distance  D  from  the 
axis  of  conductor  B  and  infinity.  This  assumes  that  the  current 
IA  in  conductor  A  is  concentrated  in  its  axis.  The  flux 
linkages  with  the  current  iA  due  to  this  flux  are  (Fig.  Ill) 


f 

JD 


log,  (13) 


per  centimeter  length  of  conductor  A. 

The  flux  due  to  the  current  iB  which  lies  at  a  distance  less 
than  D  from  the  axis  of  conductor  B  does  not  link  the  current  in 
conductor  A  and  therefore  cannot  produce  mutual-induction 
on  conductor  A, 


REACTANCE  OF  A  TRANSMISSION  LINE  359 

The  total  resultant  flux  linkages  with  the  current  iA  in 
conductor  A  per  centimeter  length  of  conductor  A  are 

2iA2  log,  y  +  ^iA2  +  2iAiB  log,  -^  (14) 

Since  a  single-phase  line  is  considered,  iA  must  be  equal  and 
opposite  to  iB  at  every  instant,  no  leakage  being  assumed.  There- 
fore 

iA  =  —IB  =  i 
and  the  expression  for  the  resultant  linkages  becomes 

2i2  log, H  &•*"  (15) 

per  centimeter  length  of  conductor  A. 

Putting  the  current  i  equal  to  unity  and  multiplying  by  10~9 
gives  for  the  resultant  inductance  in  henrys  per  conductor  per 
centimeter  length  of  line 

L  =  (2  log,  y  +  Q  10~9  henrys.  (16) 

In  terms  of  common  logarithms,  the  inductance  per  conductor 
per  centimeter  length  of  line  is 

L  =  (4.605  logio  y  +  0.5/*')  10~9  henrys.  (17) 

For  conductors  of  non-magnetic  material  such  as  are  generally 
used  for  transmission  lines,  //  is  unity. 

For  a  single-phase  transmission  line,  with  copper  or  aluminum 
conductors  or  with  conductors  of  any  other  non-magnetic  ma- 
terial, the  reactance  per  conductor  per  1000  feet  is 

ZIOGO  ft.  =  27r/  (140.4  logio  y  +  15.2)  10~6  ohms.         (18) 
The  reactance  per  mile  per  conductor  is 

-r^. 

xper  mile  =  2*f  (741  logio  -r+  80)  10-6  ohms.  (19) 

It  is  important  to  observe  that  for  a  single-phase  transmission 
line  the  reactance  per  conductor  results  from  two  parts:  the 
self-inductance  of  the  conductor  considered  and  the  mutual 
inductance  due  to  the  other  conductor.  For  a  polyphase  line, 
the  reactance  per  conductor  also  results  from  two  factors :  the  self- 
inductance  of  the  conductor  considered  and  the  mutual-induc- 
tance due  to  the  other  conductors. 


360  PRINCIPLES  OF  ALTERNATING  CURRENTS 

For  a  single-phase  line,  the  flux  due  to  the  current  iA  which 
links  conductor  B  is  equal  to  the  flux  due  to  the  current  iB 
which  links  conductor  A.  Since  the  two  currents,  iA  and  iB, 
are  not  only  equal  but  opposite  in  direction,  the  combined 
flux  linkages  produced  by  them  with  the  current  in  each  conductor 
are  zero.  The  resultant  inductance  per  conductor  of  a  single- 
phase  line  is  therefore  due  to  the  flux  the  current  in  the  conductor 
considered  would  alone  produce  between  its  axis  and  the  axis 
of  the  other  conductor,  assuming  that  the  current  is  concentrated 
in  the  axes  of  the  conductors. 

Average  Reactance  per  Conductor  of  a  Completely  Transposed, 
Ungrounded,  Three-phase  Transmission  Line. — The  lengths  of 
the  transposed  sections  will  be  assumed  equal.  The  average 
reactance  per  conductor  per  unit  length  of  line  is  equal  to  the 
average  flux  linkages  per  ampere  per  conductor  per  unit  length 
of  line,  due  to  both  self-  and  mutual-induction,  multiplied  by 
27r/,  where  /  is  the  frequency  of  the  line. 


ft 


r  r 

\ 


(a)  (6),  <*) 

FIG.   112. 

The  transpositions  required  for  complete  transposition  of  the 
line  are  shown  Fig.  112.  Parts  a,  b  and  c  of  this  figure  show  the 
three  positions  of  the  conductors.  The  D's  with  subscripts  1, 
2  and  3  are  the  distances  between  the  conductors,  measured 
between  centers.  A,  B  and  C  are  the  conductors. 

The  inductance  of  any  conductor  such  as  A  will  consist  of  three 
parts  : 

(a)  That  due  to  the  flux  linkages  with  A  resulting  from 
the  flux  produced  by  the  current  in  conductor  A,  i.e.,  by  the 
current  iA.  These  linkages  per  unit  current  are  its  self- 
inductance. 

(6)  That  due  to  the  flux  linkages  with  A  resulting  from  the 
flux  produced  by  the  current  in  conductor  B,  i.e.,  by  the  current 


REACTANCE  OF  A  TRANSMISSION  LINE  361 

iB.  These  linkages  per  unit  current  are  the  mutual-inductance 
of  B  on  A. 

(c)  That  due  to  the  flux  linkages  with  A  resulting  from  the 
flux  produced  by  the  current  in  conductor  C,  i.e.,  by  the  current 
ic.  These  linkages  per  unit  current  are  the  mutual-inductance 
of  C  on  A. 

For  the  arrangement  of  the  conductors  shown  in  (a)  of  Fig. 
112,  the  flux  linkages  with  the  current  iA  in  conductor  A  are  per 
centimeter  length  of  line :  (See  equations  (9),  (12)  and  (13).) 

/./.„  =  2iA*  log,  -^  +  I  *V  +  2iAiB  log,  £-  +  2iAic  log,  ^-  (20) 
But 

log,  y    =  log,    oc    +  log,  - 

1 
log,  £-     =  log,    oc    -f  log,  jy 

1 

log,  ^-     =   log,    oc     -f  log,   yr- 
i_J  3  •'-'3 

Therefore 

f.l.a  =  2iA2  log,  -  -}-  2*  A2  +  2z'Afc*  log,  p    +  2tAic  log,  -^ 

+  2iA  (tA  +  fa  +  tc)  log,  «  (21) 

Since  (iA  +  is  -f-  ic)  =  0  when  there  is  no  ground  connection 
which  carries  current 

2iA(iA  +  iB  +  ic)  log,  oc   =  0  (22) 

The  expression  given  in  equation  (22)  is  not  indeterminate  as 
might  appear  at  first  glance,  but  is  actually  zero.  The  linkages 
might  equally  well  have  been  taken  up  to  some  distance  such  as  x 
from  the  conductor  A,  instead  of  up  to  infinity.  If  this  had  been 
done,  the  term  in  question  would  have  been  2iA  (0)  log,  x,  which 
is  obviously  zero.  Increasing  x  to  infinity  would  not  change  the 
value  of  the  expression  which  would  still  be  zero. 

Since  the  last  term  in  equation  (21)  is  equal  to  zero 

f.l.a  =  2iA*  log,  -  +  ^2  +  2***«  lo&  5"  +  2i*ic  lo&  D~  (23) 


362  PRINCIPLES  OF  ALTERNATING  CURRENTS 

For  the  arrangement  of  the  conductors  shown  in  (b)  of  Fig. 
112,  the  flux  linkages  with  current  iA  in  conductor  A  per  centi- 
meter length  of  line  are 

f.lb  =  2iA2  log,  -  +  ^A2  +  2iAiB  log,  77  +  2iAic  log,  w  (24) 
r       ^  x>/2  -L/i 

For  the  arrangement  of  the  conductors  shown  in  (c)  of  Fig. 
112,  the  flux  linkages  with  current  iA  in  conductor  A  per  centi- 
meter length  of  line  are 

f.le.  =  2iA*  log,  -r  +  \  iA2  +  2iAiB  log,  -^  +  2^'c  log,  ~  (25) 

The  average  flux  linkages  with  current  ^  in  conductor  A  per 
centimeter  length  of  line  are 

f.Lav.    =    I    (f*La   +  f.l+   +  Ac) 

=  2t^»  log,  *  +  ^V  +  |W.  log,  STxlb^ 

ge  (26) 


Since  it  is  assumed  that  the  line  is  ungrounded  and  has  no 
neutral  connection  carrying  current 

(27) 


IB  H~  ic  =    —  iA 

and 

-ii    lo 

1               1    2'   '    1                    1 

3  Al 

&  Di  X  D2  X  D3       3^c    °ge  Di  X  D2  X  A 

=  I**A2  log,  (D!  X  Z)2  X  Z)3 

(28) 

Combining  equations  (26)  and  (28)  gives  for  the  average  flux 
linkages  with  the  current  iA  in  conductor  A  per  centimeter 
length  of  line 


2  log,  ^  +  log,  VZ),2  X  D22  X  D,2  !•     (29) 

(30) 


Similar  expressions  hold  for  conductors  B  and  C. 

It  should  be  noticed  that  for  a  completely  transposed  three- 


REACTANCE  OF  A  TRANSMISSION  LINE  363 

phase  line,  the  average  flux  linkages  per  conductor  per  unit  length 
of  line  depend  only  on  the  current  in  the  conductor  considered. 
The  resultant  reactance  drop  per  conductor  per  unit  length  of  a 
completely  transposed  three-phase  line  depends  only  on  the  dis- 
tances between  the  conductors  and  the  current  in  the  conductor 
considered.  These  statements  assume  that  there  is  no  neutral 
connection  which  carries  current.  If  there  is  a  neutral  connec- 
tion which  carries  current,  equation  (30)  does  not  hold,  since 
under  these  conditions  equation  (27)  is  not  true. 

The  average  reactance  per  conductor  per  centimeter  length  of 
a  completely  transposed  three-phase  line  which  has  no  ground 
or  neutral  connection  carrying  current  is 


*.„  =  ar/i  log,  +     ,  io-»  (31) 

ohms  per  conductor  per  centimeter  length  of  line. 

Per  mile  of  line  in  terms  of  common  logarithms  this  becomes 


xn.  =  2irf  {370.6  logic  v  ^i"  X  D^  x  D*  +  80}  10~6    (32) 


ohms  per  conductor  per  mile  of  line. 

It  is  immaterial  in  what  units  the  D's  and  r  are  expressed, 
provided  they  are  expressed  in  the  same  unit. 

If  the  conductors  are  at  the  corners  of  an  equilateral  triangle, 
Dl  =  D2  =  AJ  =  D  and 

xav.  =  27r/(741  log,0  ^  +  80}  10-6  (33) 

ohms  per  conductor  per  mile  of  line. 

By  comparing  equations  (33)  and  (19),  page  359,  it  will  be 
seen  that  the  average  reactance  per  conductor  of  a  completely 
transposed  three-phase  line,  with  conductors  at  the  corners  of  an 
equilateral  triangle,  is  equal  to  the  reactance  per  conductor  of  a 
single-phase  line  of  equal  length  and  with  equal  spacing  between 
conductors. 

If  the  conductors  of  a  completely  transposed  three-phase  line 
are  at  the  corners  of  an  isosceles  triangle,  two  of  the  distances 
DI,  D2  and  D3  will  be  equal.  Two  of  the  distances  may  also  be 


364  PRINCIPLES  OF  ALTERNATING  CURRENTS 

equal  with  the  conductors  arranged  in  a  plane.     Let  DI  =  D 
D  and  let  Z>3  =  D'.     Then 


xav.  =  27T/1  370.6  log™  +  80  110"6 


=  2^370.6  log,,  x  +  80,10- 


=  27r/{  370.6  (2  logio  y  +  |  logic  5-')  +  80}10-6 

=  27r/{  741  logio  y  +  80  +  247  loglo  ^  }  10~6  (34) 

ohms  per  conductor  per  mile  of  line. 

When  the  conductors  all  lie  in  the  same  plane,  with  equal 
distances  between  the  middle  and  each  outside  conductor, 

D' 

D  =  -x-  and  equation  (34)  becomes 

xav.  =  27T/J741  logio  ~  +  80  +  247  logio  2}  lO'6 

=  27T/I741  Iog10^  4-  154}  10-6 

ohms  per  conductor  per  mile  of  line.  (35) 

When  DI  =  D2  =  D3  =  D,  i.e.,  when  the  conductors  are  at 
the  corners  of  an  equilateral  triangle,  equations  (23),  (24)  and 
(25),  pages  361  and  362,  all  reduce  to 

f.l.  =  2iA*  loge  *  +  ~  iA*  +  2iA  (iB  +  ic)  log,  ~          (36) 

Since  it  is  assumed  that  there  is  no  neutral  connection 

IA  +  IB  +  ic  =  0 
or 

—IA  =  is  +  ic  (37) 

Combining  equations  (36)  and  (37) 

f.l.  =  2zV  log,  j  +  \  iA*  (38) 

When  the  conductors  are  at  the  corners  of  an  equilateral 
triangle,  the  flux  linkages  per  conductor  depend  only  on  the  cur- 
rent in  the  conductor  considered  and  the  distance  D  between 
conductors.  The  flux  linkages  per  conductor  and  therefore 
the  reactance  per  conductor  of  a  three-phase  line,  with  conductors 
at  the  corners  of  an  equilateral  triangle,  are  independent  of  the 


REACTANCE  OF  A  TRANSMISSION  LINE  365 

currents  in  the  conductors  other  than  the  one  considered.  This 
is  true  whether  the  load  carried  by  the  line  is  balanced  or  un- 
balanced, provided  there  is  no  neutral  connection  which  carries 
current.  Transposition  is  not  necessary  to  maintain  balanced 
line  drops  with  balanced  load.  Transposition  is  necessary, 
however,  to  prevent  mutual  induction  with  other  transmission 
lines  or  adjacent  telephone  lines. 

Problem  Illustrating  the  Calculation  of  the  Reactance  per 
Conductor  of  a  Completely  Transposed  Three-phase  Trans- 
mission Line.  —  A  certain  110,000-volt,  three-phase,  60-cycle, 
transmission  line  has  its  conductors  arranged  in  a  horizontal 
plane  with  10  feet  between  the  middle  conductor  and  each 
outside  conductor.  The  conductors  have  a  diameter  of  0.46 
inch.  What  is  the  reactance  per  conductor  per  mile  of  line? 

From  equation  (35),  page  364 

10  V  1  2 
xav.  =  2  X  3.142  X  60(741  Iog10  +  154)  10~6 


=  377  (741  X  2.72  +  154}  lO"6 

=  0.82  ohm  per  conductor  per  mile  of  line. 

Transfer  of  Power  among  the  Conductors  of  a  Three-phase 
Transmission  Line.  —  There  is  a  transfer  of  power  among  the 
conductors  of  a  three-phase  transmission  line  except  when  the 
conductors  are  at  the  corners  of  an  equilateral  triangle.  For 
simplicity,  consider  a  three-phase  line  with  conductors  at  the 
corners  of  an  isosceles  triangle.  Refer  to  equations  (23),  (24) 
and  (25),  pages  361  and  362.  Let  Dl  =  D2  =  D  and  let  D3  =  D'. 
Under  these  conditions,  if  the  instantaneous  values  of  the  currents 
are  replaced  by  their  vector  values,  the  equations  become 

f.l.a  =  21  A*  log,  -r  +  \IS  +  2~IA  (IB  log,  ^  +  Ic  log,  ^-,) 

=  21  A*  log,  y  +  ^L2  +  2  IA  Ic  log,  ^>  (39) 

/.Z.t  =  27  A2  log,  I  +  ^IA*  -\~2IA  (IB  +  Ic)  log,  ^ 

=  2/A2  log,  y  +\IA*  (40) 

f.l.c  =  21  A*  log,  *  +  \IA*  +  2~IA  (IB  log,  ^>  +  /clog,  -^) 

=  27  A*  log,  j  +  \IA2  +  2  IAIB  log,  ^7  (41) 


366  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Let  the  vectors  IA,  //*  and  Ic,  Fig.  113,  represent  the  currents 
in  the  conductors  A,  B  and  C  respectively.  Assume  clock- 
wise rotation  and  balanced  conditions  for  the  currents.  Refer  to 
equations  (39)  and  (41). 

Let  Dr  be  greater  than  D.     Under  this   condition,   log,  jy 

is  negative.  For  the  relative  position  of  the  conductors  shown 
in  Fig.  112(a),  page  360,  there  is  a  reactance  drop,  IAx  = 

2irfIA  (2  loge h  o)  >  see  equation  (39) ,  in  conductor  A  due  to 

the  current  JA.  This  reactance  drop  is  90  degrees  ahead  of  the 
current  JA  and  therefore  represents  no  power  with  respect  to 

/A.  Since  the  logarithm  of  the  ratio  jy  *s  negative  for  the 
assumed  relative  magnitudes  of  D  and  D',  the  drop,  Icx'  = 
'2irflc  (2  loge  rv)  j  see  equation  (39),  in  conductor  A  due  to  the 

current  7c  in  conductor  C  is  90  degrees  behind  Ic  as  shown.  It 
has  components  in  phase  and  in  quadrature  with  the  current 
IA,  Fig.  113.  The  effect  of  the  quadra- 
ture component  is  to  increase  the  ap- 
parent reactance  drop  in  conductor  A. 
The  effect  of  the  in-phase  component  is 
equivalent  to  an  apparent  increase  in 
the  resistance  drop  in  conductor  A. 
This  apparent  increase  in  the  resistance 
drop  in  conductor  A  does  not  represent 

FIG.   113.  ic  i 

a  loss  of  power.  It  merely  represents  a 

transfer  of  power  from  conductor  A  to  conductor  C  by  mutual- 
induction. 

For  the  relative  position  of  the  conductors  shown  in  Fig.  112(6), 
page  360,  there  is  no  mutual-induction  between  conductors 
B  and  A  or  between  conductors  C  and  A.  See  equation  (40). 
There  is  therefore  no  transfer  of  power  between  conductors  B 
and  A  or  between  conductors  C  and  A. 

For  the  relative  position  of  the  conductors  shown  in  Fig.  112(c), 

page  360,  there  is  a  reactance  drop,  JAx  =  27rfIA  (2  log, ~  +  o) » 

see  equation  (41),  in  conductor  A  due  to  the  current  IA. 
This  reactance  drop  is  90  degrees  ahead  of  the  current  7A  and 


REACTANCE  OF  A  TRANSMISSION  LINE  367 

therefore  represents  no  power  with  respect  to  JA.     Since  the 
logarithm  of  the  ratio  jy  is  negative  for  the  assumed  relative 

magnitudes  of  D  and  D',  the  drop,  JBx'  =  2*fIB(2  loge^/j  ,  see 


equation  (41),  in  conductor  A  due  to  the  current  IB  in  conductor 
B  is  90  degrees  behind  JB  as  shown.  It  has  components  both 
opposite  in  phase  to  and  in  quadrature  with  the  current  1A, 
Fig.  113.  The  effect  of  the  quadrature  component  is  to  increase 
the  apparent  reactance  drop  in  conductor  A.  The  effect  of  the 
component  which  is  opposite  in  phase  to  the  current  IA  is  to 
produce  an  apparent  decrease  in  the  resistance  drop  in  con- 
ductor A.  This  apparent  decrease  in  resistance  drop  in  the 
conductor  A  represents  a  transfer  of  power  from  conductor  B 
to  conductor  A  by  mutual-induction. 

In  a  properly  transposed  line,  the  resultant  transfer  of  power 
among  conductors  is  zero  for  the  line  as  a  whole. 

Mutual-induction  between  Transmission  Lines  or  between  a 
Transmission  Line  and  a  Telephone  Line.  —  The  mutual-induction 
between  a  transmission  line  and  a  telephone  line,  or  another 
transmission  line  which  is  parallel  to  the  first  line,  may  be  found 
in  exactly  the  same  way  as  the  mutual-induction  between  the 
conductors  of  a  single  transmission  line. 

Consider  the  mutual-induction  between  the  conductors  of  a 
three-phase  transmission  line  and  one  conductor  of  a  telephone 


FIG.  114. 


line  which  is  parallel  to  it.  Let  the  currents  in  the  transmission 
line  be  iA,  iB  and  ic  and  let  i  be  the  current  in  one  conductor  of 
the  telephone  line.  Let  DA,  DB  and  Dc  be  the  distances  in 
centimeters  between  the  conductor  of  the  telephone  line  and 
the  conductors  of  the  transmission  line  which  carry  the  currents 
zA,  IB  and  ic  respectively.  Refer  to  Fig.  114. 

The  total  flux  linkages  produced  by  the  currents  iA,  iB  and 
ic  with  the  current  i  in  one  conductor  of  the  telephone  line  per 


368  PRINCIPLES  OF  ALTERNATING  CURRENTS 

centimeter  length  of  the  telephone  line  are,  from  equation  (13), 
page  358 


f.l.  =  2iAi   log,  -£-  +  2iBi  log,      -  +  2ici  log, 
DA  UB 

=  2i   {  (IA  +  IB  +  ic) 


C 


+  iA  log,  -=r-  +  IB  log,  jr-  +  ic  log,  -=-        (42) 

-L/A  ^B  .L/c-  J 

But  (iA  +  fc's  +  ic)  =  0  since  there  is  no  neutral  connection 
which  carries  current.  Therefore 

f.l.  =  2i  {  iA  log,  ^  +  iB  log,  ^-  +  ic  log,  ^-}  (43) 

.L/A  L)B  DC  } 

The  voltage  induced  in  abvolts  per  centimeter  length  of  the 
telephone  line  is 

Vabvoli,  =  47T/  JA  log,  ~-  +  7B  log,yr-  +  7c  log,  yp  (44) 

DA  UB  DC  J 

where  IA,  IB  and  Jc  are  the  vector  expressions  for  the  currents 
in  the  conductors  of  the  transmission  line. 

If  the  portion  of  the  transmission  line  which  is  parallel  to  the 
telephone  line  be  completely  transposed,  the  resultant  flux 
linkages  with  the  conductor  of  the  telephone  line  will  be  zero, 
provided  the  transmission  line  has  no  neutral  connection  which 
carries  current.  If  there  be  a  neutral  connection  which  carries 
current,  the  relation  (iA  +  iB  +  ic)  =  0  does  not  hold. 

With  complete  transposition,  each  conductor  of  the  three- 
phase  transmission  line  must  occupy  successively  the  positions 
1,  2  and  3,  Fig.  114.  Each  position  must  be  maintained  for  one- 
third  of  the  distance  over  which  the  complete  transposition  is 
made.  The  conductors  must  be  given  a  complete  rotation  in 
position. 

The  average  flux  linkages,  produced  by  the  transposed  trans- 
mission line  are,  per  centimeter  length  of  the  telephone  line, 

2f  /.  ,        1  1  1 

f.l.av.  =  ~3[^A  log,  _  +  ^B  log,  -g-  +  ^c  loge  ^- 

+  (ic  log,  ^  +  iA  log,  A-  +  iB  log,  ^- 

+  (IB  log,  -^-  +  ic  log,  -p-  +  iA  log,  g-j  }       (45) 


REACTANCE  OF  A  TRANSMISSION  LINE  369 

Since  for  an  ungrounded  transmission  line  with  no  neutral 
connection,  (iA  +  i  B  +  ic)  =  0 

O 


-X    log.        +  iog.       +  iog.-O  (46) 

Therefore,  with  complete  transposition,  the  average  mutual- 
induction  between  an  ungrounded,  three-phase  transmission  line 
and  each  conductor  of  a  parallel  telephone  line  is  zero.  The 
resultant  voltage  induced  in  each  conductor  of  the  telephone  line 
therefore  must  be  zero.  The  above  statement  is  not  limited  to  a 
telephone  line  but  holds  equally  for  the  mutual-induction  between 
a  completely  transposed  three-phase  transmission  line  with  no 
neutral  connection  and  any  other  parallel  conductor.  The 
parallel  conductor  may  be  one  of  the  conductors  of  another 
transmission  line. 

The  average  mutual-induction  between  a  three-phase  power 
line  and  a  telephone  line  may  likewise  be  neutralized  by  trans- 
posing the  conductors  of  the  telephone  line.  If  the  conductors 
of  the  telephone  line  are  transposed,  the  transmission  line  not 
being  transposed,  the  voltage  induced  in  adjacent  transposed 
sections  of  the  telephone  line,  by  the  mutual-induction  of  the 
transmission  line,  will  be  opposite  in  phase.  If  the  two  lines 
are  parallel  and  the  transposed  sections  of  the  telephone  line  are 
equal  in  length,  these  voltages  will  be  equal  in  magnitude,  and 
in  phase  opposition,  and  their  resultant,  therefore,  will  be  zero. 
This  is  true  without  regard  to  the  relative  magnitudes  of  the 
currents,  iA,  IB  and  ic,  carried  by  the  transmission  line.  Their 
sum  may  or  may  not  be  zero.  There  may  or  may  not  be  a 
neutral  connection  which  carries  current.  Therefore,  transpos- 
ing the  telephone  line  will  eliminate  the  effect  of  mutual-induction, 
due  to  a  three-phase  transmission  line,  even  when  the  trans- 
mission line  has  a  ground  connection  which  carries  current. 

Transposing  the  telephone  line  gives  each  of  its  conductors 
equal  exposure  to  each  of  the  conductors  of  the  transmission  line 
as  well  as  to  any  neutral  connection  which  may  exist  and  must, 
therefore,  make  the  resultant  voltage  induced  in  each  conductor 
of  the  telephone  line  zero  over  any  completely  transposed  length 
of  telephone  line. 

24 


370  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Although  it  is  possible  to  neutralize  the  mutual-induction 
between  a  transmission  line  and  a  telephone  line  by  proper 
transposition  of  the  lines,  in  practice  complete  neutralization  of 
the  mutual-induction  is  often  difficult.  In  many  cases  the 
telephone  line  is  not  exactly  parallel  to  the  transmission  line. 
Under  such  conditions  the  transposed  sections  of  the  lines  are  at 
different  distances  from  each  other,  and  complete  neutralization 
of  the  mutual-induction  cannot  be  obtained  unless  the  lengths 
of  the  transposed  sections  are  varied  or  are  made  very  short, 
neither  of  which  is  practicable.  Moreover,  in  certain  cases 
taps  may  be  taken  from  the  transmission  line  at  certain  points, 
making  the  currents  unequal  in  the  different  transposed  sections. 

Voltage  Induced  in  a  Telephone  Circuit  by  a  Three-phase 
Transmission  Line  which  is  Parallel  to  the  Telephone  Line. — It 
is  not  necessary  to  find  the  linkages  with  each  conductor  of 
the  telephone  line.  All  that  is  required  is  the  flux  linkages  with 
the  loop  formed  by  the  two  conductors  of  the  telephone  circuit. 

Refer  to  Fig.  114,  page  367.  Let  the  currents  which  are  there 
marked  iA,  iB  and  ic  be  read  as  IA,  IB  and  7C  respectively,  i.e., 
let  them  be  vectors.  In  the  figure,  the  distances  from  the 
conductors  1,  2  and  3  of  the  three-phase  transmission  line  to  one 
conductor  i  of  the  telephone  line  are  DA,  DB  and  Dc.  Let  one  of 
the  unlettered  circles  be  the  other  conductor  of  the  telephone  line. 
Let  the  distances  between  this  latter  conductor  and  the  three  con- 
ductors of  the  three-phase  line  be  DA',  Db'  and  Dcr. 

The  total  linkages  with  the  telephone  circuit  per  centimeter 
length  of  telephone  line,  due  to  the  currents  JA,  IB  and  /<,•  in  the 
conductors  of  the  transmission  line,  are 

F.L.  (circuit)  =  21 A  log,  -yf  +  2IB  log,  -yf-  +  2IC  log,  -=p-      (47) 

DA  DB  Dc 

The  voltage  induced  by  these  flux  linkages  per  unit  length 
of  telephone  circuit  is 

V  =  27r/(F.L.) 

f-  CD/")2  (D  '}2  (D  '}2\ 

=  2tf  IA  log.  Vr^V  +  /.  log,  Tff4  +  Ic  log.  Trfw        (48) 


If  the  currents  in  equation  (48)  are  expressed  in  abamperes, 
the  voltage  will  be  in  abvolts.  If  the  currents  are  expressed  in 
amperes  and  equation  (48)  is  multiplied  by  10~9,  the  voltage  will 
be  in  volts. 


REACTANCE  OF  A  TRANSMISSION  LINE  371 

The  voltage  induced  per  mile  of  telephone  circuit,  using  common 
logarithms,  is 

V  =  2.33/{  IA  logio  4jf^7  +  IB  log™  4^4 

+  ic  logio  my^}  10~3  volts-    (49) 

The  currents  in  equation  (49)  must  be  expressed  in  amperes 
and  in  their  complex  form. 

It  is  convenient  in  equations  (48)  and  (49)  to  put  the  2;s, 
which  are  coefficients  of  the  logarithms  in  equation  (47),  as 
exponents  of  the  D's.  This  simplifies  the  use  of  the  equations. 
In  most  cases  in  practice,  the  distances  between  the  conductors 
of  adjacent  transmission  and  telephone  lines  are  not  directly 
known,  but  they  can  easily  be  found  from  the  known  horizontal 
and  vertical  spacing  of  the  conductors  and  the  height  of  poles 
and  distance  between  poles  of  the  two  lines.  Putting  the  2;s  in 
the  equations  as  exponents  of  the  D's,  instead  of  as  coefficients 
of  the  logarithms,  is  convenient  as  it  avoids  taking  the  square 
roots  which  would  be  necessary  in  finding  the  distances  between 
the  conductors.  The  square  of  the  distances  between  the  con- 
ductors is  readily  found  by  taking  the  sum  of  the  squares  of  the 
horizontal  and  vertical  distances  between  them. 

It  must  not  be  forgotten  that  equations  (47),  (48)  and  (49) 
are  vector  equations  and  that  all  terms  in  them  must  be  added 
vectorially.  The  currents  should  be  expressed  in  their  complex 
form,  referred  to  some  conveniently  chosen  axis. 

Voltage  Induced  in  a  Telephone  Circuit  by  an  Adjacent  Three- 
phase  Transmission  Line  which  Carries  an  Unbalanced  Load. — 
Equation  (49)  holds  for  either  balanced  or  unbalanced  loads. 
However,  when  a  careful  study  or  analysis  of  the  inductive 
effects  of  a  transmission  line  on  a  telephone  line  is  to  be  made, 
it  is  best  to  resolve  the  unbalanced  load  currents  carried  by  the 
transmission  line  into  direct-phase,  reverse-phase  and  uniphase 
components.  (See  Chapter  XII.)  The  uniphase  components, 
when  they  exist,  cause  by  far  the  most  trouble,  since  they  are  all 
in  phase.  They  are  usually  called  the  " residuals"  by  telephone 
engineers.  Since  the  residuals  are  all  in  phase,  no  amount  of 
transposition  of  the  transmission  line  will  diminish  their  inductive 
effect  on  an  adjacent  telephone  line.  To  get  rid  of  the  inductive 


372 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


effect  of  the  residuals,  the  telephone  line  itself  must  be  transposed. 
Since  the  vector  sum  of  the  direct-phase  components  and  also  the 
vector  sum  of  the  reverse-phase  components  are  zero,  they  would 
produce  no  inductive  effect  on  an  adjacent  telephone  line  if  the 
conductors  of  the  transmission  line  could  be  equidistant  from 

, 5ft ,    each  conductor  of  the  telephone  line.     This  state- 

_g[ g    ment  would  not  apply  to  the  residuals,  since  they 

are  in  phase. 

Jj  Example    of   the    Calculation    of  the   Voltage 

Induced  in  a  Telephone  Line  by  Electromagnetic 
— o    Induction. — Two  three-phase,  60-cycle,  transmis- 


1. 


sion  lines,  a,  b,  c  and  a',  &',  c',  and  a  telephone  line 
d,  e  are  carried  on  the  same  poles  with  spacings 
between   conductors   shown   in   Fig.    115.      The 
conductors  of  the  transmission  line  are  No.  000 
wire  and  have  a  diameter  of  0.41   inch.      The 
transmission  lines  are  paralleled  at  the  generating 
station,  with  corresponding  conductors  connected 
to  the  same  phase.     The   phase  order  is  a,  6,  c 
with  the  current  in  phase  a  leading  the  current  in 
phase  b.     Find  the  voltage  induced  electromag- 
netically  in   the   telephone  line,  per  mile,  when 
FIG.  115.         eacn  °f  ^ne  three-phase  transmission  lines  carries 
a  balanced  load  with  200  amperes  per  conductor. 
The  power-factors  of  lines  a,  b,  c  and  a',  b',  c'  are  unity  and 
0.85  (lagging  current)  respectively.      Also  calculate  the  voltage 
induced   in  the  telephone  line  when  each  transmission  line  is 
operated  with  an  inductive  load  of  0.80  power-factor  and  200 
amperes  per  conductor. 

The  square  of  the  distances  between  the  conductors  of  the 
transmission  lines  and  the  conductors  of  the  telephone  line  are 
given  below. 


(bd)2  =  (b'eY 

(bey  =  (Vdy 

=  (c'e)2 

=  (c'd)2 


(22)2  +  (3.5)2  =  496.3  feet  squared. 
5)2  =  486.3  feet  squared. 
5)2  =  268.3  feet  squared. 
5)2  =  258.3  feet  squared. 
S)2  =  112.3  feet  SqUared. 
5)2  =  102.3  feet  squared. 


(22)2  +  (1 
(16)2  +  (3 
(16)2  -f  (1 
(10)2 
(lO)2 


+  (S 
+  (1 


REACTANCE  OF  A  TRANSMISSION  LINE  373 

Let  (p  with  subscripts  a,  6,  c  and  a',  &',  c'  be  the  flux  linking 
the  telephone  line,  per  centimeter  length  of  telephone  line,  due  to 
the  currents  in  the  conductors  a,  6,  c  and  a',  b',  cr  respectively. 
The  magnitudes  of  these  fluxes  are: 


log*    —      =  20  X  2.303  X  0.00884  =0.407 

CL€/ 


log.  =  20  X  2.303  X  0.0165  =  0.760 

2  =  20  X  2'303  X  °-0405  =  1-866 


The  resultant  flux  through  the  telephone  circuit  per  unit  length 
of  line  is  equal  to  the  vector  sum  of  the  component  fluxes  <pa, 
(pb  and  (pc  minus  (vectorially)  the  vector  sum  of  the  component 
fluxes  <pa>,  w  and  (pc>. 

<PQ    —    <f>a    —   <f>a'    ~\-  <pb    —   <pb'    +  <f>c    ~   <pc'  (50) 

(paf  lags  (pa,  w  lags  (pb  and  <pc>  lags  (pc  by  cos"1  0.85  =  31.8 
degrees.  (pa  leads  ^  by  120  degrees  and  leads  <pc  by  240  degrees. 
(paf  leads  (pbf  by  120  degrees  and  leads  (pc>  by  240  degrees. 

Use  the  flux  (pa  as  an  axis  of  reference.     Then  the  resultant 
flux  linking  the  telephone  circuit,  per  centimeter  length  of  the 
telephone  circuit,  is 
^o  =  pa(cos  0°  -  j  sin  0°)  -  ^0<(cos  31?8  -  j  sin  31?8) 

+  <^(cos  120°  -  j  sin  120°)  -  *v(cos  151?8  -  j  sin  151?8) 
+  ^c(cos  240°  -  j  sin  240°)  -  *v(cos  271°8  -  j  sin  271°8) 
=  0.407  -j  0.000  -  0.346  +J  0.214  -  0.380  -;  0.658 

+  0.670  +j  0.359  -0.934  +j  1.616  -0.059  -j  1.865 
=  -0.642  -j  0.334 
<?o=  \/(0.642)2  +  (0.334)2 
=  0.724  lines. 

The  flux  linkages  per  centimeter  length  of  telephone  line 
are  equal  to  the  flux  per  centimeter  length  of  line  since  the  tele- 
phone line  is  a  circuit  of  one  turn. 

The  voltage  induced  in  the  telephone  circuit  per  mile  of  tele- 
phone line  is 

Vpermiie  =  27r/(0.724)  10-8  X  2.54  X  12  X  5280 

=  0.439  volts  per  mile. 

It  is  obvious  from  equation  (50)  that  if  the  loads  carried  by  the 
transmission  lines  are  both  balanced  and  have  the  same  power- 


374  PRINCIPLES  OF  ALTERNATING  CURRENTS 

factors,  the  resultant  flux  linkages  with  the  telephone  line  will 
be  zero.  Under  this  condition  the  voltage  induced  in  the  tele- 
phone line  will  also  be  zero.  Therefore,  when  both  transmission 
lines  operate  with  balanced  inductive  loads  of  0.80  power-factor 
and  200  amperes  per  line,  the  voltage  induced  in  the  telephone 
line  will  be  zero. 

Third -harmonic  Voltage  Induced  in  a  Telephone  Line  which  is 
Adjacent  to  a  Transmission  Line  that  is  Fed  from  a  Bank  of 
Transformers  which  are  Connected  in  Wye  on  Both  Primary 
and  Secondary  Sides. — The  exciting  current  in  any  transformer 
with  an  iron  core  is  not  sinusoidal  even  when  the  impressed 
voltage  is  sinusoidal.  Due  to  the  variation  in  the  permeability 
of  the  core  during  a  cycle,  the  exciting  current  will  always  con- 
tain harmonics,  among  which  the  third  is  very  prominent.  This 
third-harmonic  component  of  the  exciting  current  may  be  as 
large  as  40  or  50  per  cent,  of  the  fundamental  of  the  no-load 
current. 

The  third-harmonics  in  a  balanced  three-phase  system  are  all 
in  phase.  (See  Chapter  X.)  If  the  primary  and  secondary 
windings  of  a  bank  of  transformers  are  both  connected  in  wye, 
with  the  neutral  on  the  primary  sides  connected  to  the  source 
of  power,  the  necessary  third  harmonics  in  the  exciting  currents 
of  the  transformers  have  a  common  return  path  to  the  source 
of  power  over  the  neutral  connection.  Since  these  harmonics 
are  in  phase,  they  add  directly  on  the  neutral.  They  cannot 
exist  if  the  neutral  connection  is  opened. 

When  no  neutral  connection  is  used  on  the  primary  side,  there 
can  be  no  third-harmonic  in  the  exciting  currents.  There  will 
consequently  be  marked  third-harmonics  in  the  fluxes  of  the 
transformers  and  in  the  voltages  induced  in  the  primary  and 
secondary  windings.  These  third-harmonic  voltages  cancel 
between  the  line  terminals.  (See  Chapter  X.)  They  exist 
in  the  phase  voltages  between  the  line  terminals  and  neutral. 

If  the  neutral  on  the  secondary  side  of  the  transformers  is 
grounded,  the  third  harmonics  in  the  phase  voltages,  i.e.,  the 
voltages  to  neutral,  will  cause  a  third  harmonic  charging  current 
in  each  conductor  of  the  transmission  line  due  to  its  capacitance 
to  ground.  These  charging  currents  will  be  in  phase.  Conse- 
quently their  inductive  effects  on  an  adjacent  telephone  line  will 


REACTANCE  OF  A  TRANSMISSION  LINE  375 

be  directly  additive.  The  fundamental  and  all  harmonics  in  the 
transmission  line,  except  the  third  harmonic  and  its  multiples, 
differ  by  120  degrees  in  time  phase.  Their  inductive  effects  on 
any  adjacent  telephone  line  will  therefore  tend  to  neutralize. 
Their  resultant  inductive  effect  would  be  zero  if  the  conductors 
of  the  adjacent  telephone  line  could  be  equidistant  from  each  of 
the  conductors  of  the  transmission  line. 

Due  to  the  fact  that  the  inductive  effects  produced  by  the 
third  harmonics  add  directly,  and  also  because  of  their  triple 
frequency,  the  voltage  induced  by  them  in  an  adjacent  telephone 
line  will  be  much  greater  than  the  voltage  induced  by  fundamental 
currents  of  like  magnitude.  A  given  amount  of  third-harmonic 
voltage  in  a  telephone  line  will  produce  much  more  interference 
with  the  transmission  of  speech  than  an  equal  amount  of  funda- 
mental voltage,  since  the  third-harmonic  frequency  approaches 
that  of  the  voice  waves. 

What  has  been  said  regarding  the  third  harmonic  applies  to 
any  multiple  of  the  third  harmonic,  i.e.,  the  ninth,  fifteenth,  etc. 
The  multiples  of  the  third-harmonic  in  the  exciting  currents  of 
transformers,  however,  are  as  a  rule  small  compared  with  the 
third  harmonic  itself.  The  presence  of  ninth,  or  more  especially 
fifteenth  harmonics  would  cause  very  serious  interference  with 
telephonic  transmission  of  speech,  since  their  frequencies  are 
close  to  the  resonant  frequency  of  the  telephone  receiver,  which 
is  about  870  cycles  per  second. 


CHAPTER  XIV 

CAPACITANCE  OF  A  TRANSMISSION  LINE 

Capacitance  of  Two  Straight  Parallel  Conductors. — Consider 
two  straight,  parallel,  metallic  filaments  A  and  B  at  a  distance  d 
centimeters  apart  in  a  uniform  medium  of  dielectric  constant  K. 
Assume  that  the  filaments  are  of  zero  radius.  Let  the  charges 
per  centimeter  length  of  A  and  B  be  respectively  -\-q  and  —  q 
electrostatic  units  of  electricity.  For  such  a  system  of  charges, 
the  distribution  of  field  intensity  and  the  distribution  of  potential 
are  the  same  in  all  planes  perpendicular  to  the  axes  of  the  filaments. 
Hence,  in  treating  the  electrostatic  problems  which  arise  from  such 
a  distribution,  it  is  sufficient  to  consider  the  relations  existing  in 
one  such  plane.  This  is  the  method  which  will  be  followed  in 
this  chapter. 


FIG.   116. 


The  field  intensity  3C  at  a  point  distant  r  centimeters  from  a 
straight  filament  of  infinite  length  charged  with  +q  electrostatic 
units  of  electricity  per  centimeter  length  is 


3C  =  j=  X  —  =  —  -7-  dynes  per  unit  charge 


(1) 


The  algebraic  sign  of  the  field  intensity  is  determined  by  the 
direction  in  which  a  free  positive  charge  would  move  if  placed  in 
the  field.  A  free  positive  charge  tends  to  move  in  the  direction 
of  diminishing  potential.  The  field  intensity  3C  therefore  is 

376 


CAPACITANCE  OF  A  TRANSMISSION  LINE  377 


positive    in    a  direction  away  from  the  charge    -\-q.     This  is 
indicated  by  the  minus  sign  before-i-,  which  is  negative,  since  v 

is  decreasing  in  the  direction  of  increasing  r. 

It  should  be  remembered  in  what  follows  that  the  potential 
difference  between  two  points  is  measured  by  the  work  done  in 
moving  a  unit  positive  charge  from  one  point  to  the  other.  Po- 
tential, therefore,  in  its  fundamental  conception  cannot  be  a 
space  vector.  If  its  magnitude  varies  periodically  with  time,  as 
in  alternating-current  circuits,  it  may  then  be  considered  as  a 
time  vector. 

Let  P  be  any  point  distant  rA  and  rB  from  the  filaments  A 
and  B  respectively.  Let  R  be  a  point  equidistant  from  the 
filaments  and  at  a  distance  a  centimeters  from  each.  The 
difference  in  potential  between  the  points  R  and  P  due  to 
the  charged  filament  A  is 


-i 

KJr 


~       -    dr 


where  VPR  is  the  potential  difference  between  the  points  R  and  P 
due  to  the  charge  -\-q  per  unit  length  of  A,  and  is  measured  by 
the  work  done  in  carrying  a  unit  positive  charge  from  R  to  P. 
If  VPR  is  the  potential  difference  between  these  points  due  to  the 
charge  —  q  per  unit  length  of  B,  then 


K 


Since  the  resultant  potential  difference  between  any  two  points, 
due  to  any  number  of  charges,  is  equal  to  the  algebraic  sum  of  the 
potential  differences  due  to  each  charge  separately,  the  difference 
in  potential  between  the  points  R  and  P  must  be 

VPR  +  VPR'   =  V 

(4) 


378 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


If  the  ratio  —  =  k  is  constant,  equation  (4)  is  the  equation  of 

an  equipotential  line  and  is  a  circle.  In  any  plane  perpendicular  to 
the  axes  of  the  filaments,  the  equipotential  lines  are  circles.  The 
actual  equipotential  surfaces  due  to  the  filaments  are,  of  course, 
circular  cylinders.  That  the  equipotential  lines  are  circles  may 
be  shown  as  follows.  Refer  the  point  P,  Fig.  116,  to  rectangular 
coordinates,  taking  the  axis  of  X  and  Y  along,  and  at  right-angles 
to,  the  line  joining  A  and  B.  Let  the  origin  be  midway  between 
the  points  A  and  B,  at  b.  Then 


rv 


id         \ 

(a  -  *) 


=  k2  =  constant 


(5) 


+ 


-  A;2)  =  0 
(6) 


Equation  (6)   is  the  equation  of  a  circle.     Completing     the 
square,  gives 

Od     /l+k2\        /d\2/l+fc2\2  /d\2  22 

x  +  V  (i^F2)  +  (2)  (I^F)  +  y2  =  (2) 

/C2\  I2.  /d\2f  /I   +/C2\2 


(8) 


2 


is  the  radius  of  the  circle. 


1  - 


is  the  distance  of  the  center  of  the  circle  from  the  origin  b.     It 


CAPACITANCE  OF  A  TRANSMISSION  LINE  379 

is  to  be  noted  that  the  center  of  the  circle  lies  on  the  axis  of  X 
(Fig.  116).     The  center  lies  at  a  distance 

' 


. 

from  A  or  at  a  distance 

BA  +  AO  =  -<*- 


.     .. 

from  B. 

The  equipotential  lines  in  any  plane  perpendicular  to  the 
axes  of  the  filaments  are  two  groups  of  circles,  one  surrounding 
Ay  the  other  surrounding  B,  with  their  centers  at  a  distance 

TA 

to  the  left  of  A  when  —  =  k<l  and  at  a  distance 

rB 


/      &2 

d(-  -- 
\l  — 


i       ,J 


to  the  right  of  B  when  ^  =  k>l.     Hence  the  equi- 

potential lines  in  any  plane  perpendicular  to  the  axes  of  the  fila- 
ments are  circles  whose  centers  lie  on  the  axis  of  X,  i.e.,  on  a  line 
through  the  axes  of  the  filaments.  The  actual  equipotential 
surfaces  due  to  the  filaments  are  circular  cylinders  whose  axes 
are  parallel  to  the  axes  of  the  filaments. 

When  —  =  1,  the  radius  of  the  equipotential  line  becomes 

infinite,  equation  (8),  and  the  line  is  a  straight  line  passing 
through  b  perpendicular  to  the  line  joining  the  axes  of  the  fila- 
ments. The  corresponding  equipotential  surface  is  an  infinite 
plane  passing  through  b  and  perpendicular  to  the  line  joining 
the  axes  of  the  filaments. 

Since  the  surface  of  any  conductor  must  be  equipotential,  a 
conducting  cylinder  may  be  introduced  in  the  field  due  to  the 
charges  on  A  and  B  with  its  surface  coincident  with  any  equi- 
potential surface,  without  altering  the  field.  If  the  equipotential 
surface  in  question  is  one  of  the  group  which  surrounds  A,  the 
charge  on  the  filament  A  may  be  considered  as  existing  on  the 
surface  of  such  a  cylinder.  As  the  surface  of  any  conducting 
body  must  be  equipotential,  the  charge  will  distribute  itself  in 
such  a  way  .as  to  make  the  surface  of  the  cylinder  equipotential. 


380 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


The  effect  at  any  point  outside  of  the  cylinder  will  be  the  same 
as  if  the  charge  were  concentrated  along  the  filament  A.  It 
follows  that  any  two  parallel,  straight  conductors  of  circular 
cross-section  having  equal  and  opposite  charges  per  unit  length 
may  be  replaced  by  two  charged  filaments  at  the  points  A  and 
B,  Fig.  116,  page  376,  provided  the  charges  are  not  influenced  by 
the  presence  of  any  other  charges.  The  positions  of  the  points 
A  and  B  are  determined  by  the  radii  of  the  cylinders  and  their 
distance  apart  in  accordance  with  the  expressions  given  in  equa- 
tions (8)  to  (11)  inclusive. 

The    equipotential    surfaces    due    to    two    parallel,  straight, 
charged,  conducting,  circular  cylinders  are  shown  in  Fig.  117. 


f 


FIG.   117. 

The  solid  lines  in  this  figure  are  the  intersections  of  the  equi- 
potential surfaces  with  a  plane  perpendicular  to  the  two  charged 
cylinders.  The  solid  heavy  circles  are  the  projections  of  the 
cylinders  on  this  plane.  The  dots  in  the  circles  represent  the 
projections  of  the  filaments.  The  lines  of  electrostatic  force  due 
to  the  charges  on  the  cylinders  are  shown  dotted.  These 
lines  must  be  everywhere  perpendicular  to  the  corresponding 
equipotential  surfaces.  It  may  be  shown  that  the  lines  of  force 
are  also  circles. 


CAPACITANCE  OF  A  TRANSMISSION  LINE  381 

Remembering  that  the  two  cylinders  have  equal  and  opposite 
charges  per  unit  length,  it  follows  from  equation  (4),  page  377 
that  the  difference  in  potential  between  B  and  A  is 

TB' 


where  rAf  and  rB'  are  the  distances  for  the  cylinder  B  correspond- 
ing to  the  distances  rA  and  rB  for  the  cylinder  A  .  (See  Fig.  1  16, 
page  376.) 

Since  the  capacitance  of  any  two  conducting  bodies,  which  are 
removed  from  the  influence  of  all  other  charged  bodies,  is  equal 
to  the  ratio  of  the  charge  on  either  to  the  difference  of  potential 
between  them,  the  capacitance  of  two  parallel  straight  cylinders 
or  conductors  of  circular  cross  section,  which  are  removed  from 
all  other  charged  bodies,  must  be 


c  = 


v     2q       TBTA' 


(13) 


2  lo< 


C  is  the  capacitance  in  electrostatic  units  per  centimeter  length 
of  the  conducting  cylinders.  Equation  (13)  is  not  in  a  convenient 
form  for  use  since  rA,  rB,  rA  and  rB'  are  all  variables.  These 
may  be  replaced  in  terms  of  the  radii,  ra  and  r^  of  the  cylinders 
and  the  distance  between  their  centers. 

The  radii  of  the  cylinders  are  given  by  equation  (8),  page  378, 

d 


radii 


1  - 


This  is  positive  when  —  =  k  is  less  than  unity.     It  is  negative 
when   -  -  =  k  is  greater  than  unity.     For  the  equipotential  sur- 

' B 

face  corresponding  to  the  outer  surface  of  the  conductor  surround- 
ing A ,  —  =  k  is  less  than  unity  and  the  expression  for  its  radius 

*  B 

is  positive.     Therefore,  if  ra  is  the  radius  of  the  conductor," 

d 


382  PRINCIPLES  OF  ALTERNATING  CURRENTS 


and  rA  d  d     2 


For  the  equipotential  surface  corresponding  to  the  outer  surface 
of  the  conductor  surrounding  B,  —  -,  =  k\  is  greater  than  unity 

•  B 

and  the  expression  for  its  radius  is  negative.     Therefore,  if  rb  is 
the  radius  of  the  conductor, 

d 


and  7         rAr       d 

fc'=^=27> 

Substituting  in  equation  (13)  the  values  of  —  and  —   from 
equations  (14)  and  (15)  respectively  gives  for  the  capacitance 

K_ 

i         /   .         . —          — \  /        .          . — \     (16) 


Equation  (16)  gives  the  capacitance  in  electrostatic  units  per 
centimeter  length  of  line.  Since  d,  ra  and  rb  enter  only  in  ratios, 
it  is  immaterial  in  what  units  they  are  expressed,  provided  they 
are  all  expressed  in  the  same  unit. 

The  distance,  d,  between  the  filaments  may  be  easily  found  in 
terms  of  the  radii,  ra  and  rb,  of  the  conductors  and  the  distance, 
D,  between  their  centers.  Refer  to  Fig.  116,  page  376.  Since 

-  is  constant  and  P'Pc  is  a  circle 
rB 


r_A 
TB 

Ac  +  AP1 

Ac       AP' 
"  c£  ~  P7/? 
Ac  -  AP' 

cB  +  P'B 

2ra 

cJ5  -  P'B 
20  A 

2ra 
OA 

2ra  +  2P'B 

ra 

ra  +  d  -  (r«  -  OA) 

ra 
=  d(OA)  +  ( 

OA-    -     +  +  (17) 


CAPACITANCE  OF  A  TRANSMISSION  LINE  383 

OA  is  the  offset  of  the  filament  A  from  the  center  of  the  con- 
ductor about  A. 

The  offset,  O'B,  of  the  filament  B  from  the  center  of  the  con- 
ductor about  B  may  be  found  in  a  similar  manner.  It  is 


O'B  =  -    +      rb*+  (18) 

Then 

D  =  d  +  OA  +  O'B 


Solving  equation  (19)  for  d  gives 


,2  _  £>4  +  (ra2  -  TV*)*  -  2Z>2(ra2  -f  r62) 


-  2(ra2  -f  r>2)  (20) 


Equation  (20)  may  be  put  in  the  following  form 

dz  =  (D  +  ra  -f  rb)  (D  +  ra  —  rb)  (D  —  ra  +  rb)  (D  —  ra  —  rb)   .     . 

By  calculating  the  numerical  value  of  d  from  either  equation 
(20)  or  (21)  and  substituting  this  and  the  values  of  the  radii  in 
equation  (16),  the  capacitance  of  two  parallel,  straight  conductors 
of  circular  cross-section  may  be  found  in  electrostatic  units  per 
centimeter  length  of  the  conductors. 

Equations  (16)  and  (21)  may  be  combined  and  reduced  to  the 
following  form 

C  =  l 


2  loft  («  +  V 

= 1_ 

2  cosh-1  a 
Where 

Z)2  -ra2  -  r62 


(23) 

(24) 
2  rarb 

When  the  distance,  D,  between  the  conductors  is  great  com- 
pared with  their  radii,  ra  and  rb,  d  will  be  very  nearly  equal  to  D. 
(See  equation  (20).) 

The  conductors  of  a  transmission  line  are  of  equal  radius  and 
are  in  air.  Therefore,  for  a  transmission  line  K  =  1  and  ra  = 


384 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


rb  =  r.     When  the  radii  of  the  conductors  are  equal,  d,   from 
equation  (21),  is 

d  =  \/(D  +  2r)(D  -  2rj  = 


Substituting  this  value  of  d  and  ra  =  n  =  r  in  equation  (16), 
page  382, 


2  log. 


*+w 


1 


4  cosh~ 


D 

2r 


(25) 


(26) 


Since,   for  transmission  lines,   D  is  large   compared   with   r, 

//D\2~  /D  \ 

V  \2r/     "~     W^  ke  very  nearly  equal  to  I  «- )  and  may  be  assumed 
equal,  especially  as  it  enters  in  a  logarithm.     Putting 

D 


in  equation  (25), 


C  = 


I 


4  log,  — 


(27) 


for  the  capacitance  of  a  single-phase  transmission  line  in  electro- 
static units  per  centimeter  length  of  line. 

It  is  convenient  to  reduce  equation  (27)  to  electromagnetic 
units,  either  per  thousand  feet  or  per  mile  of  line.  It  is  also 
convenient  to  have  the  capacitance  expressed  in  terms  of  common 
logarithms.  From  the  dimensions  of  the  electrostatic  and  electro- 
magnetic units  of  capacitance  it  may  easily  be  shown  that  the 
ratio  of  the  electromagnetic  unit  of  capacitance  to  the  electrosta- 


CAPACITANCE  OF  A  TRANSMISSION  LINE  385 

tic  unit  of  capacitance  is  equal  to  the  square  of  a  velocity.  This 
velocity  has  been  shown  to  be  the  velocity  of  light  or  3  X  1010 
centimeters  per  second.  Therefore,  to  reduce  a  given  capacit- 
ance expressed  in  electrostatic  units  to  its  equivalent  capacitance 
in  electromagnetic  units,  it  is  necessary  to  divide  by  the  square 
of  the  velocity  of  light  or  by  (3  X  1010)2. 

Hence,  for  a  single-phase  line  consisting  of  two  equal,  parallel, 
straight  conductors  of  circular  cross-section,  to  get  the  capaci- 
tance in  microfarads  per  1000  feet  of  line  and  in  terms  of  common 
logarithms,  multiply  the  capacitance  given  by  equation  (27)  by 

X  (2.540  X  12  X  1000)  X 


(3  X  1010)2  2.303 

X  (109  X  106)  =  14.706  X  10-3 
r  14.7  X  10-3  _  3.68  X  10~3  (t>  . 

^microfarads  J)  J) 

4logio—  logio- 

microfarads  per  1000  feet  of  line.     Per  mile  of  line  this  becomes 

19-4  X  10-3 


microfarads 


TN 


microfarads  per  mile  of  line. 

Equation  (29)  would  be  exact  if  the  surface  densities  of  the 
charges  on  the  cylinders  were  uniform.  In  this  case  the  positions 
of  the  filaments  on  which  the  charges  may  be  considered  con- 
centrated would  coincide  with  the  axes  of  the  cylinders.  The 
distribution  of  the  charges  on  coaxial  cylinders  would  be  uniform. 

In  all  cases  occurring  in  practice,  where  it  is  necessary  to 
determine  the  capacitance  or  the  charging  current  for  a  trans- 
mission line,  the  distances  between  the  conductors  and  between 
the  conductors  and  the  earth  are  so  great  compared  with  the 
diameters  of  the  conductors  that  the  distribution  of  the  charges 
on  the  conductors  may  be  assumed  to  be  uninfluenced  except 
by  the  shape  of  the  conductors  themselves.  Since  the  conductors 
of  transmission  lines  are  nearly  always  circular  in  cross  section, 
the  charges  on  the  conductors  of  transmission  lines  may  be 
assumed  to  be  concentrated  on  the  axes  of  the  conductors. 
Equation  (29)  may  be  applied  in  the  case  of  all  aerial  transmis- 
sion lines  without  sensible  error. 

25 


386  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Example. — A  certain  two-wire  transmission  line  has  the  con- 
ductors  spaced   ten  feet   on  centers.     The  conductors  have  a 
radius   of   0.285   inch.     What   is   the    capacitance  of  the  line 
between  conductors  in  microfarads  per  mile  of  line? 
r  _  19.4  X  IP"3 
D 

logio- 

19.4  X  IP"3 
10  X  12 
logl°  -0285" 
=  19.4  X  10~3 

2.6243 
=  7.39  microfarads  per  mile. 

Capacitance  between  the  Earth  and  a  Straight  Conductor 
Parallel  to  Its  Surface. — The  capacitance  between  the  earth  and 
a  straight  conductor  parallel  to  its  surface  may  be  calculated 
on  the  assumption  that  the  earth  is  an  equipotential  surface 
of  zero  potential.  Since  the  earth  is  a  conductor,  the  electro- 
static lines  of  force  must  enter  it  perpendicular  to  its  surface. 
It  can  be  shown  that  the  electrostatic  field  between  an  infinite 
conducting  plane  and  a  charged  straight  conductor  which  is 
parallel  to  this  plane  is  the  same  as  would  be  produced  between 
one  of  two  parallel,  straight,  charged  conductors  and  the  equipo- 
tential surface  midway  between  them.  The  surface  of  the  earth 
may  be  assumed  to  be  the  equipotential  surface  ab,  Fig.  117,  page 
380.  According  to  this  assumption,  the  capacitance  between  the 
earth  and  a  conductor  parallel  to  it  at  a  perpendicular  distance  h 
above  its  surface  may  be  calculated  by  assuming  a  hypothetical 
conductor  or  image  parallel  to  the  actual  conductor  and  situated 
at  a  distance  h  below  the  surface  of  the  earth.  This  hypothetical 
conductor  or  image  will  have  a  charge  equal  and  opposite  to  the 
charge  on  the  actual  conductor.  The  charge  which  is  actually  on 
the  surface  of  the  earth  is  assumed  to  be  on  the  hypothetical 
conductor  or  image. 

Since  the  surface  of  the  earth  corresponds  to  the  equipotential 
surface  a&,  Fig.  117,  between  the  actual  conductor  A  and  its 
image  B,  the  difference  of  potential  between  the  earth  and  the 
actual  conductor  will  be  one-half  as  great  as  that  between  the 

conductor  and  its  image.     Therefore  since  C  =  -,  the    capaci- 


CAPACITANCE  OF  A  TRANSMISSION  LINE  387 

tance  of  the  conductor  with  respect  to  the  earth  will  be  twice  that 
given  by  equations  (28)  and  (29),  page  385,  or 

7  36  X  10~3 
Cto  earth  =  -  -  microfarads  per  1000  feet.        (30) 

,  Zifl 

logio  — 

38.8  X  10~3      .      . 
=  -        —  ^r  —  microfarads  per  mile.  (31) 


I 

log 


io 


When  applying  the  formulas  for  capacitance  to  earth,  it  should 
be  remembered  that  the  earth  in  the  neighborhood  of  the  con- 
ductor is  assumed  to  be  a  level  plane  of  perfectly  conducting 
material.  The  assumed  distribution  of  the  electric  field  and 
therefore  the  calculation  of  the  capacitance  may  be  considerably 
modified  by  the  presence  of  buildings,  trees  and  poles,  especially 
steel  poles.  It  will  also  be  influenced  by  the  variation  in  the 
height  of  the  conductor  above  the  ground,  as  well  as  by  the  fact 
that  the  earth,  in  certain  localities  or  in  dry  weather,  may  differ 
considerably  from  a  perfect  conductor.  It  has  been  shown 
experimentally  that  the  equipotential  plane  corresponding  to  the 
surface  of  the  earth  actually  lies  below  the  earth's  surface. 
Experiments  carried  out  at  the  Massachusetts  Institute  of 
Technology  in  Cambridge  have  shown  it  to  be  approximately 
coincident  with  the  mean  tide-water  level. 

Charging  Current  of  a  Transmission  Line.  —  Consider  a  three- 
phase  transmission  line  with  all  three  conductors  in  a  single  plane 
which  is  parallel  to  the  surface  of  the  earth.  Assume  the  surface 
of  the  earth  to  be  a  perfectly  conducting  plane.  According  to 
this  assumption  the  image  of  any  conductor,  as  for  example 
conductor  1,  Fig.  118,  will  be  at  a  distance  below  the  surface  of 
the  earth  equal  to  the  height,  h,  of  the  conductor  above  the 
surface.  A  line  joining  any  conductor  and  its  image  must  be  per- 
pendicular to  the  surface  of  the  earth,  since  this  surface  forms  the 
equipotential  plane  of  zero  potential.  (See  Fig.  117,  page  380.) 

Let  the  distances  between  the  middle  conductor  and  each  of 
the  outside  conductors  be  equal.  Call  this  distance  D.  Let  h 
be  the  height  of  the  conductors  above  the  earth.  Assume  the 
distances  between  the  conductors  and  also  the  height  of  the 
conductors  above  the  earth  to  be  large  compared  with  the  radii. 
Under  these  conditions  the  charges  on  the  conductors  may  be 


388 


PRINCIPLES  OF  ALTERNATING  CURRENTS 


considered  as  concentrated  along  the  axes  of  the  conductors,  so 
far  as  any  effect  outside  of  the  conductors  is  concerned.  Figure 
118  shows  the  conductors  of  the  line  and  their  images. 

All  charges  and  voltages  will  be  expressed  as  vectors.  Let  the 
charges  on  the  conductors  per  centimeter  length  of  line  be  Qit 
Q2  and  Q3  electrostatic  units.  The  corresponding  charges  on  the 
images  will  then  be  —  Qi,  —Q?  and  —  Q3.  Let  the  radii  of  the 
conductors  be  r. 


•*•(&--  C 

!                          < 

k--^ 

fe. 

1 

1 

1  1 

h   \<  D  ^ 
kd-^ 

<  D  > 

•--f/c-B 

h' 

h' 

!                 i     Ea, 

th     | 

i  i 

////////////////// 

////////// 

i  i 

i  ' 

i 

i  : 

1   ! 

i  :  _ 

_ 

1'  2'  3' 

FIG.   118. 

Consider  first  the  potential  difference,  Fi2,  between  conductors 
1  and  2  due  to  the  three  actual  charges  on  the  conductors  and 
the  three  apparent  charges  of  opposite  sign  on  the  images.  The 
difference  of  potential  between  conductors  1  and  2  is  equal  to  the 
algebraic  sum  of  the  potential  differences  which  would  be  pro- 
duced by  the  three  charges  and  the  three  images  each  acting 
separately. 

Fi2  =  2^!  log,  ^  +  2Q2  log,  ^  +'  2Q3  log,  ^ 


-Qi)  log. 


2h 


+  2(  -Q,)  log. 


2(  -Q8)  log. 


(2D) 


(32) 


CAPACITANCE  OF  A  TRANSMISSION  LINE 

Similarly, 

on  n  r 

F28  =  2Q,  log,  -p  +  2Q2  log,7  +  2Q3  log,  ^ 


389 


log,  ~  +  2Q2  log,      +  2Q3  log, 


1      (33) 


(34) 


The  computation  may  be  somewhat  simplified  by  placing  the 
2's  before  the  logarithms  as  exponents  of  the  quantities  under  the 
logarithms.  Making  this  change, 


«  =  «i  log,         X 


log.         X 


v 
4  X 


— 

=  Ci  log,  T  x 


D 


log,        x 


4/i2 


2  ^2     I 

x  -- 


__ 

+  g3iog, 


x 


(35) 

(36) 

(37) 


F3i  =  < 

Also 

Fi2  +  F23  +  Fsi  =  0  (38) 

Qi  +  Q2    +  Qs    =0  (39) 

Knowing  Fi2,  F23  and  F3i  in  complex,  Qi,  Q2  and  Q3  may  be 
found  in  complex. 

Equations  (35),  (36)  and  (37)  hold  only  for  electrostatic 
units.  The  ratio  of  the  electromagnetic  unit  of  charge  to  the 


390  PRINCIPLES  OF  ALTERNATING  CURRENTS 

electrostatic  unit  of  charge  is  equal  to  the  velocity  of  light  or 
3  X  1010  centimeters  per  second.  The  corresponding  ratio  for 
the  units  of  voltage  is  the  reciprocal  of  the  velocity  of  light  or 

10    reciprocal     centimeters     per    second.     Therefore,     if 
o  X  J-U 

volts  are  used  in  equations  (35),  (36),  (37)  and  (38),  the  resulting 
expressions  for  Qi,  Q%  and  Qs  derived  from  these  equations  must 
be  multiplied  by 


10"  X 


X  10|  =  9  X      11       (40) 


in  order  to  get  the  charges  in  coulombs. 

Inspection  of  equations  (35),  (36)  and  (37)  will  show  that  the 
effect  of  the  images  on  the  potentials  and  hence  on  the  charges 
is  negligible  for  ordinary  heights  and  spacings  of  the  conductors 
of  transmission  lines.  For  example,  consider  the  first  term  in 
equation  (35).  If  the  image  —  Qi  is  neglected,  this  term  becomes 

D2 


Suppose  D  is  10  feet  and  the  height  h  is  40  feet.  Let  the  radii 
of  the  conductors  be  0.5  inch.  Then  the  first  term  actually  is 

{100*X144X        4  X  1600 
j  —  025-     <4  x  1600  +  100 

2.303  Qi  logio  {57,600  X  0.985}  =  2.303  Qi  X  4.7537 
If  the  image  be  neglected,  the  first  term  becomes 

2.303  Q!  logiof  57,600}   =  2.303  Ql  X  4.7604 

Neglecting  the  image  in  this  case  makes  an  error  of  only  0.14 
of  one  per  cent  in  the  first  term  of  the  equation  for  Viz-  Neglect- 
ing the  images  will  produce  errors  of  the  same  order  of  magnitude 
in  the  other  terms. 

If  the  effect  of  the  images  is  neglected,  equations  (35),  (36) 
and  (37)  become 

V  12  =  Qi  log,  ~  +  Q2  log,  ~  +  Q3  log,  \  (41) 

F23  =  Qi  loge  4  +  Q2  log.  ~  +  Q3  log,  ^  (42) 

2  4D2 

-j  +  Q,3  log,  -3-  (43) 


CAPACITANCE  OF  A  TRANSMISSION  LINE  391 

The  charging  current  may  be  obtained  from  the  charges  in  the 
following  manner, 

q  =  Qm  sin  a>t 

dq     '    ~ 
i  =  -^  =  wQm  cos  wt 

...  (44) 


The  charging  currents  in  amperes  per  conductor  per  centimeter 
length  of  line  may  be  found  by  multiplying  the  root-mean-square 
values  of  the  charges  per  conductor  per  centimeter  length  of 
line  by  co  =  2irf.  If  the  drop  in  potential  along  the  line  is  negligi- 
ble, the  total  charging  current  may  be  found  by  multiplying 
the  charging  current  per  unit  length  of  line  by  the  length  of  the 
line. 

Although  the  solution  of  the  preceding  equations  for  the 
charges  on  the  conductors  of  a  three-phase  line  is  a  simple  matter 
when  the  conductors  are  in  a  plane  parallel  to  the  surface  of  the 
earth,  the  equations  do  not  take  into  account  the  effects  of  the 
poles  and  steel  towers  or  the  effect  of  adjacent  lines.  They  also 
neglect  the  effect  of  adjacent  trees  and  buildings,  irregularities 
of  the  earth's  surface,  etc.  They  assume  that  the  equipotential 
surface  due  to  the  earth  is  actually  coincident  with  the  earth's 
surface,  an  assumption  which  may  be  considerably  in  error, 
especially  in  very  dry  sections  of  the  country.  This  last  source 
of  error,  however,  is  of  little  consequence  since  the  effect  of  the 
images  on  the  charges  is  so  small  as  to  be  negligible  under  ordinary 
conditions.  An  exact  solution  taking  into  account  all  the  factors 
just  mentioned  is  obviously  impossible.  These  considerations, 
together  with  the  fact  that  with  ordinary  spacing  of  the  con- 
ductors of  a  transmission  line  and  the  height  of  the  conductors 
above  the  earth  the  charging  current  is  due  almost  entirely  to 
the  capacitance  between  the  conductors  themselves,  make  it 
possible  to  neglect  the  effect  on  the  charging  current  of  the 
earth  and  other  adjacent  bodies. 

Example  of  the  Calculation  of  the  Charging  Current  of  a 
Three-phase  Transmission  Line.  —  A  1  10,000-volt,  three-phase, 
60-cycle  transmission  line  has  conductors  1.14  inches  in  diameter, 
which  are  in  a  horizontal  plane  and  are  spaced  10  feet  between 
the  middle  and  each  outside  conductor,  at  a  height  of  40  feet 


392  PRINCIPLES  OF  ALTERNATING  CURRENTS 

above  the  earth.  Assuming  there  is  no  drop  in  voltage  along 
the  line,  find  the  charging  current  in  each  conductor  per  mile  of 
line.  The  line  will  be  assumed  not  to  be  transposed,  and  the 
capacitance  of  the  line  to  earth  will  be  neglected.  Balanced 
voltages  will  be  assumed.  The  arrangement  of  the  line  and  the 
notation  used  will  be  the  same  as  in  Fig.  118,  page  388. 
Take  Fi2  as  the  axis  of  reference.  Then 

7i2  =  110,000(1+^0) 
=  110,000  +  jO 


723=  110,000    -  *  -j 
=  -55,000  -j95,260 


=  110,000   -    + 

=  -55,000  +  j95,260 
From  equations  (41),  (42)  and  (43),  page  390 


LO,OC      f  JO   =  2.303  |  Ql  Iog10^^~--  +  £2  loglo 

1 


-55,000  -  j'95,260  =  2.303 


X  9  X  1011      (45) 

(10  X  12)2 
logio  4  +  Q2  logio  — (o^7)s~- 


+  £3  loglo  X  9  X  10U      (46) 

-55,000  +  J95,260  =  2.303 


loglo        -  X  9  X  10"      (47) 


110,000  +jO  =  96.31    X    10nQi   --    96.31    X    10nQ2 

-  12.48  X  10UQ3     (48) 
-55,000  -  j95,260  =  12.48   X   10nQi  +  96.31    X   10UQ2 

-  96.31  X  10nQ3     (49) 
-55,000  +  j95,260  =   -108.8  X  10uQi 

+  108.8  X  10UQ:}     (50) 
Qi  +  £2  +  Q3  =  0  (51) 


CAPACITANCE  OF  A   TRANSMISSION  LINE  393 

Substitute  in  equation  (51)  the  value  of  Q2  from  equation 
(48). 

--        [96.31  X  IQ11-^       12.48  X  IQ11-^  110,000     |      -- 

Vl-|~  196.31  X  10llVl     96.31  X  lO11^3     96.31  X  lO11!"^3" 
Ql  +  0.4352  Q3  -  0.571  X  10~8  =  0  (52) 

Substitute  in  equation  (50)  the  value  of  Qi  from  equation  (52). 
-55,000  +  j95,260  =  -108.8  X  10n(-0.4352Q3 

+  0.571  X  10~8)  +  108.8  X  1011  Q3 

Q3  =  45.63  X  10-11  +  J609.8  X  lO'11  coulombs  per  cm.      (53) 

Substitute  in  equation  (50)  the  value  of  Q3  from  equation  (53)  . 

-55,000  +  j'95,260  =  -108.8  X  10llQi 

+  108.8  X  10U(45.63  X  10-11  +  J609.8  X  10~n) 
Qi  =  551  X  10~n  -  J265.7  X  1Q-11  coulombs  per  cm.      (54) 

Substitute  in  equation  (51)  the  values  of  Q3  and  Qi  from  equa- 
tions (53)  and  (54)  respectively. 

551  X  10-11  -  J265.7  X  10-11  +  Q2  +  45.63  X  lO"11 

+  j'609.8  X  10-11  =  0 

Q2  =  -597  X  10-11  -  J344.1  X  1Q-11  coulombs  per  cm.    (55) 
The  numerical  values  of  the  charges  are 

Qi  =  V(551)2  +  (265.7)2  X  1Q-11 

=  612  X  10~n  coulombs  per  cm.  length  of  line. 

Q2  =  V(597)2  +  (344.  1)2  X  10~n 

=  689  X  10~u  coulombs  per  cm.  length  of  line. 

X  10~u 


=  612  X  10~n  coulombs  per  cm.  length  of  line. 
To  get  the  charging  current  in  amperes  per  mile  of  line,  the 
charges  in  coulombs  per  centimeter  length  of  line  must  be  multi- 
plied by 

27r/(2.540  X  12  X  5280)  =  377  X  160,900 

=  6.066  X  107 

/i  =  612  X  10-11  X  6.066  X  107 
=  0.371  amperes  per  mile  of  line. 

72  =  689  X  10-11  X  6.066  X  107 

=  0.418  amperes  per  mile  of  line. 

73  =  612  X  10-11  X  6.066  X  107 

=  0.371  amperes  per  mile  of  line. 


394  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  currents  7i  and  73  are  equal  since  the  conductors  1  and  3 
are  symmetrically  placed.  If  the  line  were  transposed,  as  all 
transmission  lines  are  in  practice,  the  average  charging  current 
per  mile  or  per  unit  length  of  line  would  be  the  same  for  all 
conductors.  This  assumes  balanced  voltages. 

Capacitance  of  a  Balanced  Three-phase  Transmission  Line 
with  Conductors  at  the  Corners  of  an  Equilateral  Triangle, 
Neglecting  the  Effect  of  the  Earth.  —  When  the  effect  of  the  earth 

and  adjacent   bodies   is   neglected,  the 

lQQi  expressions  for  the  charging  currents  and 

/     \  for  the  capacitance  of  a  balanced  three- 

/  \  phase  transmission  line,  with  conductors 

D  jR  at  the  corners  of  an  equilateral  triangle, 

/  \  become  very  simple.     Refer  to  Fig.  119. 

Q  /  \  2          Let    the   distances  between  the  con- 

(§£  -----  D  —     ~~o^     ductors    be    D,    assumed    to    be    large 

FIG.  no.  compared    with    the    radii.      Let    the 

charges  on  the  conductors   1,  2  and  3 

be,  respectively,  Q1;  Q2  and  Q3  electrostatic  units  per  centimeter 
length  of  conductor. 

Consider  the  conductors  1  and  2.  The  charge  on  conductor  3 
cannot  produce  any  difference  of  potential  between  conductors  1 
and  2  since  it  is  the  same  distance  from  each.  The  potential 
difference  Fi2  due  to  Q3  is 


=  0  (56) 

The  resultant  potential  difference  between  conductors  1  and  2, 
due  to  all  three  charges,  is  that  produced  by  charges  Qi  and  Q* 
only. 


-  Q2)  loge  j  (57) 

in  a  vector  sense,  as  is  indicated  in  the  equation.  For  balanced 
condition  of  impressed  voltages,  since  the  line  is  symmetrical, 
Qi  and  Q2  must  be  equal  in  magnitude  and  must  differ  by  120 
degrees  in  time  phase.  The  magnitude  of  the  vector  difference 
of  any  two  equal  vectors  which  differ  in  time  phase  by  120  degrees 
is  equal  to  the  magnitude  of  either  vector  multiplied  by  the 


CAPACITANCE  OF  A  TRANSMISSION  LINE  395 

square  root  of  three.     Hence  for  balanced  conditions,  the  alge- 
braic form  of  equation  (57)  may  be  written 

V=2V3Q  log,  y  (58) 

or 

Q  =  -  —  -r  (59) 


Equation  (59)  gives  the  magnitude  of  the  charge  Q  per  con- 
ductor per  centimeter  length  of  line  in  electrostatic  units  in 
terms  of  the  magnitude  in  electrostatic  units  of  the  voltage  V 

between  conductors.     Since  i  =  -TT,  the  charging  current  per 

conductor  per  centimeter  length  of  line  is 

2x/F 
/  =  -  —  ~.  statamperes  (60) 


where  V  is  expressed  in  statvolts,  i.e.,  in  electrostatic  units. 
Both  I  and  V  are  root-mean-square  values.  If  V  is  expressed  in 
volts,  the  right-hand  member  of  the  equation  must  be  multiplied 

by  Q       1Q11  to  give  the  current  in  amperes. 

/««™.r«.  =  —  —    JQi  —  per  conductor  per  centimeter 


•  amperes 


2-v/31og«  — 1  9  X  1011  length  of  line     (61) 

T  j 

It  will  be  seen  from  equation  (57)  that  the  voltage  Viz  is 
in  phase  with  the  vector  difference  of  the  charges  Qi  and  Qz, 
i.e.,  in  phase  with  (Qi  —  Q2).  Similarly  Vzz  is  in  phase  with 
(Q2  —  Qz)  and  Vz\  is  in  phase  with  (Q3  —  Qi).  For  balanced 
conditions,  the  three  charges  Qi}  Q2  and  Q3  must  be  equal  in 
magnitude  and  must  differ  by  120  degrees  in  time  phase.  The 
only  way  in  which  the  charges  can  be  equal  in  magnitude  and 
differ  in  phase  by  120  degrees,  and  at  the  same  time  have  their 
vector  differences  taken  in  pairs  in  phase  with  the  line  voltages, 
is  for  the  charges  to  be  in  phase  with  the  wye  voltages  of  the 
system. 

D  D 

Viz  =  VIQ  -  VZo  =  2Q!log,  —  -  2Q2Jog€y 

(62) 


396  PRINCIPLES  OF  ALTERNATING  CURRENTS 

where  the  voltages  Fio  and  Vzo  are  the  voltages  between  the 
neutral  and  conductors  1  and  2  respectively.  The  phase  rela- 
tions between  the  voltages  and  charges  will  be  made  clearer  by 
referring  to  Fig.  120.  This  figure  shows  the  phase  relations 
between  the  vectors  in  equation  (62). 

The  charges  on  the  conductors  per  unit  length  of  line  are 
the  same  as  would  exist  on  three  equal  condensers  connected  in 
wye  across  the  three-phase  line,  each  condenser  having  a  capaci- 
tance of 

-Q-. = l—    statf arads  (63) 


-^  _&e  -       2  log,  - 
V3  r  r 

Equation  (63)  gives  the  equivalent  capacitance,  in  electrostatic 
units,  per  phase  to  neutral,  per  centimeter  length  of  a  balanced 
three-phase  transmission  line,  with  con- 
ductors at  the  corners  of  an  equilateral 
triangle.     It  gives  the  capacitance  for 
each  of  three  equal  condensers   which, 
if    connected    in    wye   across   the   line, 
would  take  the  same  charging  current 
as   the   line  actually  takes.      Equation 
(63)  holds  only  for  balanced  conditions. 
It  should  be  noticed  that  the  equiva- 
FIQ.  120.  l611^  capacitance  to  neutral  of  a  balanced 

three-phase  transmission  line,  with  con- 
ductors at  the  corners  of  an  equilateral  triangle,  is  equal  to 
twice  the  capacitance  between  conductors  of  a  single-phase  line 
with  the  same  conductor  spacing.  (Equation  (27),  page  384.) 
It  is  equal  to  the  capacitance  to  neutral  of  the  single-phase  line. 
In  terms  of  common  logarithms  and  microfarads  per  mile  of 
line,  equation  (63)  becomes  (see  equation  (29),  page  385). 

r       38.8  X  10-3      . 

C  — =r —  microfarads  per  mile  of  line  (64) 

logio  - 

The  charging  current  per  conductor  per  mile  of  a  balanced 
three-phase  line  with  conductors  at  the  corners  of  an  equilateral 
triangle,  neglecting  the  drop  in  voltage  along  the  line,  is 


CAPACITANCE  OF  A  TRANSMISSION  LINE  397 

27r/Fn  X  38.8  X  IP"9 

D 

logic  - 

2  44  F  f  X  10~7 

—  r  --  •  amperes  per  conductor  per  mile  of  line.    (65) 


Vn  is  the  voltage  to  neutral.  Equations  (57)  to  (65)  inclusive 
neglect  everything  except  the  line  itself.  They  hold  only  for 
balanced  conditions. 

When  the  conductors  of  a  balanced,  transposed  transmission 
line  are  not  at  the  corners  of  an  equilateral  triangle,  the  average 
capacitance  of  the  line  to  neutral  may  be  calculated  by  using  the 
spacing  of  the  equivalent  equilateral  arrangement.  As  an 
approximation,  this  spacing  may  be  taken  equal  to 


D  =  ^D,  XD2XD3  (66) 

where  DI,  Dz  and  D&  are  the  actual  distances    between    the 
conductors. 

Example  of  the  Calculation  of  the  Average  Charging  Current 
per  Conductor  per  Mile  of  a  Transposed  Line  by  the  Use  of  the 
Equivalent  Equilateral  Spacing.  —  A  1  10,000-volt,  60-cycle, 
three-phase,  transposed  transmission  line  has  its  conductors 
arranged  in  a  horizontal  plane  40  feet  above  the  earth,  with  10 
feet  between  the  middle  conductor  and  each  outside  conductor. 
The  radius  of  the  conductors  is  0.57  inch.  What  is  the  average 
charging  current  per  conductor  per  mile  of  line.  The  equivalent 
equilateral  spacing  will  be  used  in  calculating  the  charging 
current. 

D  =  ^/D,  XDtXDs 


I  = 


=  v"lO  X  10  X  20 
=  12.60    feet. 
2.447n/  X  10-7 


- 


244X  110>OQQ  X  60  X  10-7 

V3  9.30  X  10-  X  10-7 

/  =  - 


log] 


12.60  X  12  logio  265.3 


0.57 
=  0.384  amperes  per  conductor  per  mile  of  line. 


398  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The  actual  charging  current  per  conductor  per  mile  of  the  com- 
pletely transposed  line  may  be  found  from  the  charges  per  centi- 
meter length  of  each  conductor  when  the  line  is  not  transposed. 

These  charges  were  found  to  be  (see  page  393). 

Qi  =  551  X  10-11  -  J265.7  X  lO'11 
Q2  =  -597  X  10-11  -J344.1  X  1Q-11 
Q3  =  45.63  X  10-11  +  J609.8  X  1C)-11 

coulombs  per  conductor  per  centimeter  length  of  line,  where 
Qi,  Qz  and  Q3  are  the  charges  on  the  three  conductors  with  the 
conductors  in  the  positions  shown  in  Fig.  118,  page  388. 

If  the  line  is  transposed  so  that  the  conductors  from  left  to 
right,  Fig.  118,  are  3,  1,  2,  the  charge  Qi  on  conductor  1 
would  be  the  same  in  magnitude  as  Q2  given  above,  but  it  would 
not  be  in  phase  with  it  since  the  middle  conductor  would  now  be 
connected  to  phase  1  instead  of  to  phase  2.  Since  the  cyclic 
order^of  the  voltages  used  in  finding  the  charges  Qi,  Q2  and  Qs 
was  Viz,  ^23,  Vsi  with  Vzs  lagging  Viz,  the  charge  Qi  on  conductor 
1  when  it  occupies  the  middle  position,  i.e.,  the  position  2,  Fig.  118, 
may  be  found  both  in  magnitude  and  in  time  phase  by  multi- 
plying Q2  by  the  operator  which  produces  a  rotation  of  +120 
degrees. 

If  Q  ~  Qi  is  the  charge  on  conductor  1  per  centimeter  length 
of  line  when  this  conductor  occupies  the  position  shown  in  Fig. 
118,  the  charges  on  it,  per  centimeter  length  of  line,  when  it 
occupies  the  positions  2  and  3  may  be  found  by  rotating  the 
charges  Qz  and  Q3  as  given  above  through  +120  and  —120 
degrees  respectively. 

The  charges  on  conductor  1  per  centimeter  length  of  line  when 
the  conductor  occupies  the  three  positions  necessary  for  complete 
transposition  are 

Position  1     Q  =  (551  -  j265.7)10-n 

Position  2    Q  =  (-597  -j  344.1)  (-0.500  +  J0.866)10-n 

=  (596.5  -  J344.9)10-n 
Position  3     Q  =  (45.63  +  j609.8)(-  0.500  -  j0.866)10-n 

=  (505.3  -  j344.4)10-n 


CAPACITANCE  OF  A  TRANSMISSION  LINE  399 

Average        O  =  I  {  (551  +  596.5  +  505.3)  + 
o 

j(-  265.7  -  344.9  -  344.4)  JIG"11 
=  (551  -  j318)10-u 
Average       Q  =  V(551)2  +  (318)  2  X  10~n 

=  636  X  10~u  coulombs  per  cm.  length  of  line. 

The  actual  charging  current  per  conductor  per  mile  of  the  com- 
pletely transposed  line  is 

/  =  2-nf  X  636  X  10-11  X  (2.540  X  12  X  5280) 
=  0.386  amperes 

This  differs  by  less  than  one  per  cent  from  the  value  found  by 
the  use  of  the  equivalent  equilateral  spacing. 

Voltage  Induced  in  a  Telephone  or  Telegraph  Line  by  the 
Electrostatic  Induction  of  the  Charges  on  the  Conductors  of  an 
Adjacent  Transmission  Line.  —  A  telephone  line  is  on  the  same 
poles  as  a  three-phase  transmission  line,  whose  conductors  are 
in  a  plane  parallel  to  the  surface  of  the  earth  and  at  a  distance  h 
above  it.  Let  the  distance  between  the  middle  and  each  of  the 
outside  conductors  be  D.  Let  the  two  telephone  conductors,  A 
and  B,  be  at  a  distance  2d'  apart  and  each  at  a  distance  h'  above 
the  surface  of  the  earth.  Also,  let  the  telephone  conductors  be 
at  equal  distances  from  a  line  joining  the  middle  conductor  of 
the  transmission  line  and  the  image  of  this  conductor. 

Refer  to  Fig.  118,  page  388.  Let  the  charges  on  the  conductors 
of  the  transmission  line  per  unit  length  of  line  be  Qi,  Q2  and  Q3 
electrostatic  units.  The  potential  difference  between  conductor 
A  of  the  telephone  line  and  the  earth,  due  to  the  charges  +Qi  on 
conductor  1  of  the  transmission  line  and  —  Q\  on  the  image  of 
conductor  1,  is 

VAO'  =  2Ql\og,V(h—h,)2+(D_dv 

=  statvolts      (66) 


h,)2  +  (D  _ 

Similar  expressions  hold  for  the  voltages  VAO"  and  VAO"  between 
conductor  A  and  the  earth  due  to  the  charges  on  conductors  2 
and  3  and  their  images.  The  resultant  potential  difference 


400  PRINCIPLES  OF  ALTERNATING  CURRENTS 

between  conductor  A  and  the  earth,  due  to  all  three  charges 
Qi,  Q2  and  Qz  and  their  images  —Qi,  —Qz  and  —  Q3,  is 


(h  -hfy  +  (D  - 

+  < 


+  <2*  log.  statvolts         (67) 

If  VBO  is  the  resultant  difference  of  potential  between  the 
conductor  B  and  the  earth,  found  in  the  same  manner  as 
the  potential  difference  VAO,  the  potential  difference  between  the 
conductors  A  and  B  of  the  telephone  line  is 

VAB  ==   VAO  ~   VBO  (68) 

In  order  to  evaluate  VAO  and  VBO,  it  is  necessary  first  to  calciK 
late,  by  the  method  already  given,  the  charges  Qi,  $2  and  Q3  in 
complex  from  the  voltage,  height  and  spacing  of  the  conductors 
of  the  transmission  line.  Substitute  these  values  of  the  charges 
in  equation  (67)  for  VAO  and  in  the  corresponding  equation  for 
VBO-  The  voltage  VAB  then  may  be  found  from  equation  (68). 
In  most  cases  it  is  not  necessary  to  consider  the  effect  of  the 
images. 

Transposition,  when  it  can  be  properly  carried  out  for  either 
the  power  line  or  the  telephone  line,  will  eliminate  the  resultant 
voltage  in  an  adjacent  telephone  line  due  to  eletromagnetic  in- 
duction. With  the  power  line  transposed,  the  induced  voltages 
neutralize  in  such  a  length  of  the  telephone  line  as  is  exposed  to 
a  complete  transposition  of  the  transmission  line.  With  the 
telephone  line  transposed,  the  voltages  induced  in  adjacent  sec- 
tions of  the  telephone  line  are  equal  and  opposite  and  neutralize. 
When  either  the  telephone  line  or  the  transmission  line  is  trans- 
posed, the  voltage  induced  by  electromagnetic  induction  in  any 
section  of  a  telephone  line  will  be  too  low  to  be  troublesome  or 
dangerous. 

Transposition,  when  it  is  properly  carried  out  for  either  the  tele- 
phone line  or  the  transmission  line,  will  eliminate  the  resultant 
voltage  in  a  telephone  line  due  to  electrostatic  induction  of 
an  adjacent  transmission  line,  i.e.,  it  will  make  the  resultant 
voltage  acting  along  the  line  zero.  Transposition,  however,  will 


CAPACITANCE  OF  A   TRANSMISSION  LINE 


401 


not  eliminate  the  danger  from  the  voltage  induced  by  electrostatic 
induction,  as  the  magnitude  of  this  voltage  is  independent  of  the 
exposed  length  of  the  telephone  line.  The  voltage  in  a  telephone 
line  due  to  the  electrostatic  induction  of  an  adjacent  transmission 
line,  and  especially  a  single-phase  transmission  line  with  ground 
return,  may  not  only  be  troublesome  but  also  dangerous  to  We. 
This  situation  may  be  met  by  providing  a  shunt,  of  low  reactance 
for  transmission  frequency,  between  the  conductors  of  the  tele- 
phone line  and  between  these  conductors  and  the  earth.  This 


;\ 


•P« 

<H* 


Earth 


y///////////////^ 

FIG.   121. 


arrangement  will  allow  the  induced  voltages  on  the  telephone 
line  to  be  dissipated  by  a  flow  of  current.  The  shunts,  if  properly 
designed,  will  be  of  sufficiently  high  reactance  for  telephonic 
frequencies  to  prevent  serious  leakage  of  the  telephonic  current. 

When  the  voltages  of  the  transmission  line  are  unbalanced, 
the  best  way  to  study  the  effects  of  electrostatic  induction  on  an 
adjacent  telephone  or  telegraph  line  is  to  resolve  the  voltages 
of  the  transmission  line  into  their  direct-phase,  reverse-phase  and 
Uniphase  or  residual  components.  (See  Chapt.  XII.) 

An  Example  Illustrating  the  Calculation  of  the  Voltage  Induced 
in  a  Telephone  Line  by  Electrostatic  Induction. — A  110,000-volt, 

26 


402  PRINCIPLES  OF  ALTERNATING  CURRENTS 

60-cycle,  three-phase  transmission  line  and  a  telephone  line  are 
carried  on  separated  pole  lines  along  the  same  right-of-way. 
The  spacings  of  the  conductors  of  the  transmission  line,  and  of 
the  telephone  line,  the  distance  between  the  poles,  and  the  height 
of  the  conductors  above  the  earth  are  shown  in  Fig.  121.  The 
cyclic  order  of  the  voltages  of  the  transmission  line  is  1-2-3  with 
the  voltage  2  lagging  the  voltage  1.  The  conductors  of  the  trans- 
mission line  are  No.  0000  and  have  a  diameter  of  0.46  inch.  Find 
the  electrostatic  voltage  in  volts  induced  on  the  telephone  line. 
The  images  will  be  neglected.  The  voltages  of  the  transmission 
line  will  be  assumed  to  be  balanced. 

The  squares  of  such  distances  between  conductors  as  will  be 
needed  in  the  solution  of  this  problem  are: 

(la)2  =  (40  -  25  +  12  X  0.866) 2  +  (50  -  I)2  =  3045 

(16) 2  =  (40  -  25  +  12  X  0.866) 2  +  (50  +  I)2  =  3245 

(2a)2  =  (40  -  25)2  +  (50  -  6  -  I)2  =  2074 

(26)*  =  (40  -  25)2  +  (50  -  6  +  I)2  =  2250 

(3a)2  =  (40  -  25)2  +  (50  +  6  -  I)2  =  3250 

(36)2  =  (40  -  25)2  +  (50  +  6  +  I)2  =  3474 

Since  the  conductors  of  the  transmission  line  are  at  the  corners 
of  an  equilateral  triangle,  the  charge  per  conductor  per  centimeter 
length  of  line  is  (Equation  (63),  page  396  or  Equation  (59), 
page  395) 

Q  =  -^  X  C  = — 


12  X  12      3 
2\/3  X  2.303  logio 


x'xio-* 


0.23 


=  4930  X  -    X  10-2 

o 


state oulombs  per  cm.  length  of  line. 

+ Wg       X8X 

volts,  where  the  Q's  are  expressed  in  statcoulombs. 


CAPACITANCE  OF  A   TRANSMISSION  LINE  403 

Taking  Qi  as  an  axis  of  reference, 
Va6  =  2.303  X  4930(0.02763(1  +  JO) 

+  0.03537(-0.5  -  jO.866) 

+  0.02895(-0.5  +  J0.866))  volts 
=  11,350(0.02763  +  JO  -  0.01769  -  jO.03063 

-0.01447  +  j  0.02507  j  volts 
=  11,350  X  (-0.00453  -  jO.00556)  volts 
Vat.  =  11,350  V(0.00453)2+  (0.00556) 2 
=  81.3  volts, 

Another  Example  Illustrating  the  Calculation  of  the  Voltage 
Induced  in  a  Telephone  Line  by  Electrostatic  Induction. — In  the 
preceding  problem,  calculate  the  voltage  between  conductor  a  of 
the  telephone  line  and  earth,  owing  to  the  electrostatic  induction 
of  the  transmission  line.  In  this  case  the  images  cannot  be 
neglected. 

The  square  of  the  distances  between  conductor  a  of  the  tele- 
phone line  and  each  conductor  of  the  transmission  line  and  also 
between  conductor  a  of  the  telephone  line  and  each  of  the  images 
of  the  conductors  of  the  transmission  line  will  be  needed. 

The  squares  of  these  distances  are : 

(la)2  =  (40  -  25  +  12  X  0.866)2  +  (50  -  I)2  =  3045 

(2a)2  =  (40  -  25)2  +  (50  -  6  -  I)2  =  2074 

(3a)2  =  (40  -  25)2  +  (50  +  6  -  I)2  =  3250 

(1'a)2  =  (40  +  25  -f  12  X  0.866)2  +  (50  -  I)2  =  8085 

(2'a)2  =  (40  -f  25)2  -f  (50  -  6  -  I)2  =  6074 

(3'a)2  =  (40  +  25)2  +  (50  +  6  -  I)2  =  7250 

(le)2  =  (l'e)z  =  (40  -f  12  X  0.866)2  =  2539 

(2e)2  =  (2'e)2  =  (40) 2  =  1600 

(3e)2  =  (3'e)2  =  (40)2  =  1600 

In  the  preceding  table,  1',  2'  and  3'  indicate  images  and  the 
e's  indicate  earth. 


404  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Q  =  4930  X  -Q-  X  10~2  statcoulombs  per  centimeter  length  of 
o 

line.     (From  preceding  problem.) 


+  (-Qj)  log,  X  3  X  102  volts. 

\o  Ct/     J 

where   the  Q's  are  expressed  iri  statcoulombs  per   centimeter 
length  of  line. 

8085    .    n   .       6074    .  7250 

Q2  log,  2074+^3  log, 


X  3  X  102  volts. 

Taking  Qi  as  an  axis  of  reference, 
Vae  =  2.303  X  4930  {0.4241  (1  +  jQ)  +  0.4667  (-  0.5  -  jO.866) 

0.3485  (-  0.5  +  J0.866)}  volts. 
=  11,350  X  (0.0165  -  ./0.1024)  volts. 
Vae  =  11,350\/(0.0165)2  +  (0.1024)2 
=  1173  volts. 


CHAPTER  XV 

SERIES-PARALLEL  CIRCUITS  CONTAINING  UNIFORMLY  DIS- 
TRIBUTED RESISTANCE,  REACTANCE,  CONDUCTANCE 
AND  SUSCEPTANCE 

Series-parallel  Circuits. — Series-parallel  circuits  are  discussed 
in  Chapt.  VII.  In  that  chapter,  the  circuits  considered  are 
those  whose  series  impedance  and  parallel  admittance  are  con- 
centrated between  definite  points.  The  circuits  are  lumpy 
circuits,  not  smooth  circuits  with  uniformly  distributed  constants. 

When  the  series  impedance  and  parallel  admittance  of  a 
circuit  are  uniformly  distributed,  the  equations  for  the  resultant 
current  and  voltage,  either  at  the  terminals  or  at  any  point 
along  the  line,  are,  for  steady  conditions,  comparatively  simple 
if  expressed  in  terms  of  hyperbolic  functions  of  complex  quanti- 
ties. Tables  and  charts  are  now  available  by  means  of  which 
such  functions  may  be  evaluated.1  Even  if  the  tables  are  not 
at  hand,  the  equations  may  be  solved  without  difficulty  by  con- 
verting the  hyperbolic  functions  of  complex  quantities  into 
hyperbolic  functions  of  real  quantities,  each  multiplied  by  an 
operator  of  the  form  (cos  0  +  j  sin  0). 

The  most  important  example  of  a  circuit  having  uniformly 
distributed  constants  is  an  ideal  transmission  line.  An  ideal 
transmission  line  is  one  which  has  its  conductor  resistance 
and  conductor  reactance  and  also  its  conductance  and  suscept- 
ance  between  conductors  uniformly  distributed  along  the  line. 
The  conductor  resistance  is  the  ohmic  resistance  of  the  conductor, 
corrected  for  skin  effect.  The  conductor  reactance  is  discussed 
in  Chapt.  XIII.  The  conductance  and  susceptance  between 
conductors  are  due  to  the  leakance  and  capacitance  between 
conductors.  The  capacitance  of  a  transmission  line  is  discussed 
in  Chapt.  XIV.  The  leakance  between  conductors  includes 

1  Tables  of  Complex  Hyperbolic  and  Circular  Functions,  A.  E.  Kennelly, 
Harvard  University  Press. 

405 


406  PRINCIPLES  OF  ALTERNATING  CURRENTS 

the  leakage  over  insulators  and  poles  and  the  loss  caused  by  dis- 
charge through  the  air  due  to  corona.  Under  normal  operating 
conditions,  the  leakage  over  insulators  and  poles  and  the  corona 
discharge  should  be  practically  negligible  for  a  properly  designed 
transmission  line. 

An  actual  transmission  line,  when  considered  between  points 
at  which  branches  are  taken  off,  conforms  nearly  enough  to  the 
conditions  for  a  uniform  line  to  be  treated  as  such  without 
sensible  error,  i.e.,  to  be  treated  as  a  line  whose  conductor  resis- 
tance and  reactance  per  unit  length  of  line  and  whose  conductance 
and  susceptance  between  conductors,  also  per  unit  length  of 
line,  are  constant. 

The  derivation  of  the  general  equations  for  voltage  and  current 
for  a  uniform  line  under  steady  conditions  of  operation  is  an 
easy  matter.  It  involves  merely  the  solution  of  two  simple 
simultaneous  differential  equations  for  voltage  and  current 
and  the  determination  of  the  constants  of  integration  from  the 
terminal  conditions  of  the  line. 

Equations  for  Voltage  and  Current  of  a  Transmission  Line 
whose  Series  Resistance  and  Reactance  per  Unit  Length  of  Line 
and  whose  Parallel  Conductance  and  Susceptance  per  Unit 
Length  of  Line  are  Constant. — The  equations  which  will  be  de- 
veloped hold  only  for  steady  conditions. 

Consider  a  circuit  having  uniformly  distributed  resistance, 
reactance,  conductance  and  susceptance.  Let 

r  =  resistance  per  conductor  per  unit  length  of  line. 

x  =  reactance  per  conductor  per  unit  length  of  line. 

g  =  leakage  conductance  between  conductors  per  unit  length 

of  line. 

b  =  susceptance  between  conductors  per  unit  length  of  line, 
due  to  the  capacity  effect  between  conductors. 

No  actual  transmission  line  can  be  exactly  uniform  on  account 
of  the  variation  produced  in  its  constants  by  many  factors,  such 
as  temperature,  atmospheric  pressure,  weather  conditions,  etc. 
Irregularities  in  the  spacing  of  the  conductors  will  affect  the 
reactance,  conductance  and  susceptance  of  a  line.  The  con- 
ductance and  susceptance  will  also  be  affected  to  some  extent 
by  the  irregularities  in  the  height  of  the  conductors  above  the 


SERIES-PARALLEL  CIRCUITS  407 

earth's  surface.  The  effect  of  this  variation  in  height  will  be 
small  for  an  ordinary  transmission  line.  The  weather  conditions 
affect  the  corona  loss  and  the  leakage  loss  between  conductors. 
The  weather  conditions,  therefore,  will  affect  the  conductance 
between  conductors.  If  a  section  of  a  transmission  line  crosses 
a  high  mountain  pass,  the  conductance  of  the  section  may  be 
much  increased,  especially  during  storms,  by  the  low  atmos- 
pheric pressure.  In  practice,  average  values  of  the  constants  are 
used  for  calculating  the  performance  of  transmission  lines. 

The  resistance,  r,  may  be  determined  from  the  size,  material 
and  average  temperature  of  the  conductors,  due  allowance  being 
made  for  skin  effect.  The  method  of  calculating  the  reactance, 
x,  is  given  in  Chapt.  XIII.  The  conductance,  g,  may  be  found 

by 

_  Corona  loss  +  leakage  loss  (both  per  unit  length  of  line) 

(Voltage  between  conductors)2 

The  corona  loss  may  be  calculated,  with  sufficient  accuracy, 
by  means  of  empirical  formulas.1  The  leakage  loss  may  be 
determined  from  existing  experimental  data.  For  a  well  designed 
transmission  line  under  good  operating  conditions,  both  the 
corona  loss  and  the  leakage  loss  should  be  negligible. 

The  susceptance,  b,  may  be  found  from  the  relation  6  =  wC, 
where  C  is  the  capacitance  between  conductors  per  unit  length 
of  line.  Capacitance  of  transmission  lines  is  discussed  in 
Chap.  XIV. 

All  values  of  r,  x,  g,  b,  z  and  y  must  be  given  per  unit  length  of 
line.  Although  the  unit  of  length  chosen  is  merely  a  matter  of 
convenience,  the  mile  is  commonly  used  in  practice. 

z  =  r  +  jx  (1) 

y  =  g  +  jb  (2) 

The  plus  sign  is  used  before  b  in  equation  (2)  since  b  represents 
capacitive  susceptance. 

Let  p  be  a  point  distant  L  from  one  end  of  the  line.  The 
fundamental  differential  equations  for  the  current  and  voltage 
at  the  point  p  are 

dl  =  yVdL  (3) 

dV  =  ~zl  dL  (4) 

1  Dielectric  Phenomena  in  High  Voltage  Engineering,  F.  W.  Peek,  Jr. 


408  PRINCIPLES  OF  ALTERNATING  CURRENTS 

From  which 

=  ^   •  <«> 


Equations    (5)    and    (6)    are   the   current   and   the   potential 
gradients,  respectively,  at  a  distance  L  from  one  end  of  the  line. 
Differentiating  equations  (5)  and  (6)  with  respect  to  L  gives 

d2!       -  dV 


~dT*  ~      dL 

Combining  equations  (7)  and  (6)  and  equations  (8)  and  (5) 
gives 

0  (9) 


-  yzV  =  0  (10) 


Equations  (9)  and  (10)  are  the  differential  equations  for 
current  and  voltage  of  a  transmission  line. 

Equations  (9)  and  (10)  are  linear  differential  equations  of  the 
form 


p  =  emq  is  a  solution  of  equation  (11).     The  complete  solution 
of  equation  (11)  is 

p  =  Al€miq  +  A2em'fl 
alm2emq  +  a2memq  +  asem(I  =  0 

aim2  +  a2m  +  a-6  =  0 

From  equations  (9)  and  (10)  a\  =  1,  a2  =  0  and  a3  =  —yz. 
m2  —  yz  =  0 

m  =   ±\/P 
The  solutions  of  equations  (9)  and  (10)  are,  therefore, 


(13) 


SERIES-PARALLEL  CIRCUITS  409 

Equations  (12)  and  (13)  apply  only  to  the  steady  state,  i.e., 
after  any  transient  has  disappeared. 

The  constants  of  integration,  AI,  A2,  BI  and  Bz  must  be 
determined  from  assumed  terminal  conditions  either  at  the  load 
or  at  the  source  of  power.  Let  them  be  determined  from  the 
assumed  terminal  conditions  at  the  load.  For  this  case,  L  is 
the  distance  from  the  receiving  end  of  the  line  to  the  point  at 
which  the  voltage  and  current  are  desired.  It  must  be  reckoned 
positive  when  measured  from  the  load  end  of  the  line  to  the  source 
of  power.  Let  the  assumed  current  and  voltage  at  the  receiving- 
or  load-end  of  the  line,  i.e.,  for  L  =  0,  be  IR  and  VR  respectively. 

Differentiate  equation  (12)  with  respect  to  L  and  put  L  =  0 
and  V  =  VR.  This  gives 

(£1  —     Wp  r^_      -WP  -  — 

yV 


y  VK  =  AI  \/$z  -  A2  Vp  (14) 

Putting  L  =  0  and  V  =  VR  directly  in  equation  (12), 

JR  =  A!  +  A2  (15) 

Combining  equations  (14)  and  (15), 

yVa  =  (In  -  A  2)  \/yz  -  A2\/p 
From  which 

_  IR  \/yz  —  y  VR 


(16) 

*    \  \   'A  I 

and 

(17) 


The  constants  BI  and  Bz  may  be  found  in  a  similar  manner 
from  equation  (13). 


Substituting  the  values  of  the  constants  AI,  A2,  BI  and  #2  in 


410  PRINCIPLES  OF  ALTERNATING  CURRENTS 

equations   (12)   and   (13)   gives   the   current   and   voltage   at   a 
distance  L  from  the  receiving  end  of  the  line. 


7R  coshL-v/P  +  "F*A/     sinh  Z/Vp  (22) 

,_.  . 

;+'\»2Vi  (24) 

=  yK  cosh  L\/yz  +  /a  A/-  sinh  L\/yi  (25) 

'    c/ 

7  and  F  in  equations  (22)  to  (25)  are  the  current  and  voltage 
at  a  distance  L  from  the  receiving  or  load  end  of  the  line. 


The   expressions   \/yz,  *       and   ^       are    the   fundamental 

constants  of  a  transmission  line. 

If  L  is  measured  from  the  generator  end  of  the  line,  i.e.,  is 
positive  from  the  source  of  power  towards  the  load,  equations 
(22)  and  (25)  become 


=  IG  cosh  L     P  -  VG-  sinh  £  Vp  (26) 


-  7G  J|  si 

*  c/ 


sinh  L  Vp  (27) 


where  7  and  F  are  the  current  and  voltage  at  a  distance  L  from 
the  generator  end  of  the  line.  I0  and  VG  are  the  current  and 
voltage  at  the  generator  end  of  the  line. 

.»  /-  is  called  the  surge  impedance  and  is  denoted  by  ZQ.     +   _ 

is  called  the  surge  admittance  and  is  denoted  by  yQ.  Both 
the  surge  impedance  and  the  surge  admittance  are  complex 
quantities. 


is  known  as  the  propagation  constant  and  is  denoted  by  a. 


SERIES-PARALLEL  CIRCUITS  411 


It  follows  from  Chapt.  I,  pages  21  and  22,  that 


(28) 


cos  „  (Bz  +  By)  +  j  sin  =  (B2  +  Bv)  (29) 

L  & 

=  a  \Ba 

=  «r+  j<x2  (30) 

where 

0y  =  tan"1- 

x 
r 


cos  ^  (0,  +  «,)  (31) 

sin  \  (By  +  a,)  (32) 

Also 

2  (^  -  ^)  (33) 

(34) 

^  (^  ~  ».)  (35) 

=  2/0  ^o  (36) 

If  F«  is  taken  as  the  axis  of  reference  and  BR  is  the  phase  angle 
of  the  load  current,  7R,  with  respect  to  the  voltage,  VR,  at  the 
load,  equations  (22)  and  (25)  may  be  written  in  the  following 
form. 


7  =  IR   OR  cosh  {LcL^\  +  F«0^oosmh  {La^\      (37) 
y  =  VR\Q_  cosh  {La\0«}  +  IR  \0*  zQ\^smh  {La\J*a}      (38) 

These  equations  may  be  evaluated  by  the  use  of  tables  of 
hyperbolic  functions  of  complex  quantities.  (See  foot  note, 
page  405.) 

Interpretation    of    Equations    (20)    and    (23). — Substituting 


412  PRINCIPLES  OF  ALTERNATING  CURRENTS 


ai  -f-  j«2  for  \/yz  and  y0  for  */|  in  equation  (20)  and  remember- 
ing that  eL(a>  +ia*)  may  be  written  eLa>  €jLa\ 

I  =  \  (la  +  yoV*)  eLa'  eiLa*  +  ™  (/*  -  2/o  V«)  e~La>  €-*"•        (39) 

For  increasing  values  of  L,  z.e.,  in  going  from  the  load  to  the 
generator,  the  first  term  of  the  second  member  of  equation  (39) 
is  a  vector  which  increases  in  magnitude  logarithmically  with  L, 
on  account  of  the  factor  eLa'.  This  first  term  advances  in  phase 
at  a  rate  which  is  directly  proportional  to  L,  due  to  the  factor 
J1"*.  This  term  may  be  written 


I  Direct  =      (In  +  2/0  V  R)  eL<Xl  (cos  La2  +  j  sin  La2) 


(40) 


where  ^R  is  the  admittance  of  the  load.  If  equation  (40)  is  a 
vector  which  increases  in  length  and  advances  in  phase  in  going 
from  load  to  generator,  it  may  equally  well  be  represented  by  a 
vector  which  decreases  in  length  and  retards  in  phase  in  going 
from  generator  to  load.  Hence,  the  first  term  of  the  second  mem- 
ber of  equation  (39)  represents  a  wave  travelling  from  the  gener- 
ator to  the  load. 

The  second  term  of  the  second  member  of  equation  (39)  is  a 
vector  which  decreases  in  magnitude  logarithmically  with  L,  due 
to  the  term  e~Lctl.  This  term  retards  in  phase  at  a  rate  which 
is  directly  proportional  to  L,  due  to  the  term  e~jLo(2.  This  term 
may  be  written 

I  Reflects  =  2  (!R  ~  yo'VR)  <TLot'  (rosLaa  -  j  sin  L«2) 

=  \VR(yR-yQ}e-L^\-LoL^  (41) 

7  Direct  is  a  direct  wave  traveling  from  the  generator  to  the  load. 
I  Rejieded  is  B>  wave  reflected  at  the  load  which  travels  from  the 
load  to  the  generator. 


SERIES-PARALLEL  CIRCUITS  413 

The  two  terms  of  the  second  member  of  equation  (23)  may  be 
considered  to  represent  direct  and  reflected  voltage  waves. 


(42) 


V  Reflected    =         (V  R   ~   20/ft) 


(43) 


where  ZR  is  the  impedance  of  the  load  and  7*  is  the  load  current. 

Equations  (41)  and  (43)  show  that  reflection  of  the  current 
and  voltage  waves  will  not  occur  at  the  load  when  yR  =  yQ  or 
2ft  =  20,  the  equality  being  considered  in  a  vector  sense.  In 
general,  reflection  will  always  occur  at  a  point  in  a  circuit  where 
there  is  a  change  in  the  electrical  constants  of  the  circuit. 

The  direct  and  reflected  waves  combine  to  produce  standing 
waves  with  maximum  and  minimum  points,  i.e.,  loops  and  nodes, 
equally  spaced  along  the  line.  As  these  loops  and  nodes  must 
occur  a  quarter  wave-length  apart,  they  will  not  show  on  lines 
of  present  commercial  length  and  frequency,  since  a  quarter 
wave  length,  even  for  sixty  cycles,  is  about  775  miles. 

A  more  complete  discussion  of  direct  and  reflected  waves  will 
be  found  in  works  dealing  specifically  with  long-distance  power 
transmission.1 

Propagation  Constant,  Attenuation  Constant,  Wave-length 
Constant,  Velocity  of  Propagation  and  Length  of  Line  in  Terms 
of  Wave  Length.  —  It  was  stated  on  page  410  that 


is  known  as  the  propagation  constant.     The  real  part  of  the 
propagation  constant,  i.e., 

«i  =  Vy~z  cos  ^  (By  +  0.)  (44) 

determines  the  attenuation  of  the  traveling  waves.     It  is  known 
as  the  attenuation  constant.     (Equation  (39),  page  412.) 

1  Electric  Phenomena  in  Parallel  Conductors,  F.  E.  Pernot. 


414  PRINCIPLES  OF  ALTERNATING  CURRENTS 

The   coefficient   of   the   imaginary   part   of   the   propagation 
constant,  i.e., 

«2  =  Vyz  sin  2  (By  +  0.)  (45) 

determines  the  amount  of  phase  retardation  of  the  component 
waves  in  their  direction  of  propagation.     It  is  known  as  the 
wave-length  constant.     (Equation  39,  page  412.) 
The  wave  length,  X,  of  the  traveling  waves  is 

X  =  ^  (46) 

<*2 

where  «2  is  in  radians.  The  unit  in  which  the  wave  length,  X, 
is  expressed  depends  on  the  unit  of  length  selected  for  the  con- 
stants y  and  z.  If  they  are  per  mile  of  line,  X  will  be  in  miles. 
If  they  are  in  any  other  unit,  X  will  be  in  the  same  unit.  The 
magnitude  of  the  product  yz  is  independent  of  the  units  chosen 
for  admittance  and  impedance,  provided  corresponding  units 
are  used,  i.e.,  mhos  and  ohms  or  abmhos  and  abohms. 

X=—  ^  -  (47) 

\/yz  sin  ^  (®y  +  #*) 

(Equation  (32),  page  411.) 

The  length,  L,  of  a  transmission  line,  expressed  in  terms  of  the 
wave  length  for  the  impressed  frequency,  is 


The  velocity  of  propagation  is 

X/  =  2^  (49) 

where  /  is  the  frequency. 

yz  =  (g  +  jb)(r  +  jx)  =  («i  +  jo?2)2 

(rg  —  xb)  +  j(rb  +  gx)  =  «i2  +  jcloL\a.^  —  CL<? 

«i2  —  «22  —  rg  —  xb 

2aiai  =  gx  +  rb 

( 


SERIES-PARALLEL  CIRCUITS  415 

Solving  for  a\  and  0% 

*y  ~  x6  +  iv)  (50) 


(*</  +  *6  -  rg)  (51) 

If  there  were  no  transmission  losses,  r  and  0  would  each  be 
zero.  Under  this  condition,  there  would  be  no  attenuation  and 
the  wave  length  constant  would  be 

0.1  =  \/xb 

=  o>\/LC  (52) 

where  co  =  27r/  and  L  and  C  are  the  inductance  and  capacitance 
per  unit  length  of  line. 

For  a  line  without  transmission  losses,  the  wave  length  would 
be 

2ir          27r  1  ,KO, 

X  =  —  =  -    -7-=  —  -  (53) 


From  the  dimensions  of  inductance  and  capacitance,  it  may  be 
shown  that  /=-^  has  the  dimensions  of  velocity.  This  velocity 

has  been  shown  to  be  the  velocity  of  light,  or  3  X  1010  centimeters 
per  second.  It  is  the  limiting  velocity  of  propagation  for  electric 
waves.  The  actual  velocity  of  propagation  is  slightly  less  than 
this  on  account  of  the  terms  zy  and  —  rg  in  the  expression  for  «2 
(Equation  (51)).  The  difference  between  the  actual  velocity 
and  that  found  by  assuming  no  transmission  losses  is  not  great 
as  a  rule. 

Solution  of  the  Transmission  Line.  —  If  a  table  of  Hyperbolic 
Functions  of  Complex  Quantities  is  available,  equations  (22) 
and  (25)  or  (37)_and  (38),  pages  410,  411Lshould_be  used  for  cal- 
culating IG  and  Vo  at  the  generator  when  7«and  Va  at  the  load  are 
known.  For  calculating  IR  and  VR  at  the  load  when  70  and  Vo  at 
the  generator  are  known,  equations  (26)  and  (27)  should  be  used. 

The  open-circuit  voltage  at  the  load  is  found  by  putting  IR 
equal  to  zero  in  equation  (25),  page  410,  or  in  equation  (38), 
page  411  and  then  solving  for  VR. 

The  charging  current  of  the  line,  i.e.,  the  no-load  current,  may 
be  found  for  fixed  generator  voltage  by  substituting,  in  equation 


416  PRINCIPLES  OF  ALTERNATING  CURRENTS      4 

(22),  page  410,  or  in  equation  (37),  page  411,  the  no-load  volt- 
age at  the  receiving  or  load  end  of  the  line,  as  found  above,  and 
then  solving  for  7G,  IR  being  zero. 

The  length  of  the  line  in  terms  of  the  wave  length  of  the 
impressed  voltage  is  found  from  equation  (48),  page  414. 

The  efficiency  of  transmission  is 

PR  =   VR!R  COS  BR 

PG     VGIG  cos  eG 

where  PR  and  PG  are  the  power  at  the  load-end  and  the  generator- 
end  of  the  line  respectively.  The  angles  0R  and  6G  are  the 
power-factor  angles  at  the  load-end  and  the  generator-end  of  the 
line  respectively.  The  power  may  be  found  as  indicated,  but  it 
may  be  more  convenient  to  find  it  from  the  complex  expressions 
for  current  and  voltage. 

Evaluation  of  the  Equations  for  Current  and  Voltage  when 
Tables  of  Hyperbolic  Functions  of  Complex  Quantities  are  not 
Available.  —  From  equation  (30),  page  411. 

cosh  La  =  coshL(«i  +  ja2) 

_   1 


(cos  La2  +  j  sin  L«2) 
- 


+  -e~LQr'  (cos  Laz  —  j  sin  La2) 


+  j\(eLa>  -e-1""')  sin  L«2 
—  cosh  La\  cos  La:2  +  j  sinh  La\  sin  Loti       (54) 

It  may  be  shown,  in  a  similar  manner,  that 

sinh  La  =  sinhLai  cos  La2  +  j  cosh  Lai  sin  La2  (55) 

Then 

Vo  =  VR  cosh  La  +  7#2o  sinh  La 

=  Fftfcosh  Lai  cos  L«2  +  j  sinh  Lai  sin  La2\ 

Lai  cos  La2  +  j  cosh  Lai  sin  La2[     (56) 


SERIES-PARALLEL  CIRCUITS  417 

where 

zo  =  20    cos (^  -  ev)  +j sin(02  -  0 


IG  =  IR  cosh  La  +  VRyo  sinh  La 

=  IR(  cosh  Lai  cos  La2  +  j  sinh  Lai  sin  La2 j 

+  F«2/o{sinh  Lai  cos  La2  +  j  cosh  Lai  sin  La2}      (57) 

where 

1  1 

cos  ^(By  —  ez)  +  j  sm  -(ev  —  Bz)  j 

I 

COS  ^(By  +   Bz} 

1. 

=  Vyz  sin  ^(By  +  9.) 


Fo,  F/z,  /o  and  7a  must  be  in  complex.  If  VR  and  /«  are 
known,  the  solution  of  equations  (56)  and  (57)  will  give  VG  and  JG 
in  complex.  If  VG  and  IG  are  known,  the  solution  of  these 
equations  will  give  VR  and  IR  in  complex. 

The  open-circuit  voltage  at  the  load  or  receiving  end  of  the 
line  is  found  by  putting  7B  equal  to  zero  in  equation  (56). 

F*(open  circuit)  =  -         = ^V      -y r-^—      (58) 

cosh  Lai  cos  La2  +  J  sinh  Lai  sin  La2 

When  JR  is  zero,  the  first  term  of  the  second  member  of 
equation  (57)  is  zero.  If  VR'  is  the  open-circuit  voltage  at  the 
load  end  of  the  line,  found  from  equation  (58),  the  charging  cur- 
rent is  given  by 

J0(charging)  =  VR'y0  {cos  \(BV  -  0.)  +jsin^(0v  -  0.)  1 

L  & 

X  {sinh  Lai  cos  La2  +  j  cosh  Lai  sin  La2 j      (59) 

When  a  transmission  line  is  not  uniform,  due  to  loads  being 
taken  off  at  intermediate  points  or  to  a  change  in  conductor 
spacing  or  in  size  of  conductors,  it  must  be  considered  in  sections 
of  such  length  that  the  principles  of  uniform  lines  are  applicable. 
The  ends  of  the  sections  will  be  at  the  points  where  loads  are 
applied  or  where  the  conductor  spacing  or  the  size  of  conductors 
changes. 
27 


418  PRINCIPLES  OF  ALTERNATING  CURRENTS 

Example  of  the  Calculation  of  the  Performance  of  a  Trans- 
mission Line  for  Steady  Operating  Conditions  from  its  Electrical 
Constants.— A  150-mile,  25-cycle,  110,000-volt,  3-phase  trans- 
mission line  has  its  conductors  placed  at  the  corners  of  an  equi- 
lateral triangle.  The  conductors  are  10  feet  apart  on  centers. 
Each  conductor  is  a  19-strand  copper  cable  with  an  overall 
diameter  of  %  inch  and  an  equivalent  cross  section  of  300,000 
circular  mils.  The  skin  effect  at  25  cycles  for  conductors  of  the 
size  used  for  this  line  is  very  small  and  may  be  neglected.  The 
line  operates  below  its  corona  voltage,  i.e.,  below  the  voltage  at 
which  the  loss  due  to  corona  discharge  between  conductors 
appears.  The  line  insulation  is  assumed  high.  The  corona  loss 
and  the  leakage  between  conductors  over  line  insulators,  etc. 
will  be  assumed  to  be  negligible.  Calculations  will  be  based  on 
an  average  line  temperature  of  20  degrees  centigrade. 

What  are  the  voltage,  current,  power  and  power-factor  at  the 
power  station  or  generator  end  of  the  line  for  an  inductive  load 
at  the  receiving  end  of  30,000  kilowatts  at  110,000  volts  and 
0.90  power-factor.  What  is  the  efficiency  of  transmission? 

If  the  voltage  at  the  generator  end  of  the  line  remains  constant 
at  the  value  required  for  the  given  load  conditions,  what  will  be 
the  charging  current  of  the  line  and  the  open-circuit  voltage  at 
the  receiving  or  load  end  of  the  line? 

What  is  the  length  of  the  line  in  terms  of  the  wave  length  of 
the  current  and  voltage  waves? 

The  resistance  of  the  copper  conductors  will  be  assumed  to  be 
10.58  ohms  per  mil-foot  at  20  degrees  centigrade.  The  mile 
will  be  used  as  the  unit  of  length. 

coon 

r  =  10.58  X  ^~  v"A  =  0.1862  ohm  per  mile. 
jUUU 


g  =  negligible  (assumed  zero). 

38.8  X  10-3       38.8  X 
C  (to  neutral)  =  - 


,         10  X  12  2.584 

logio     5/ 

71 Q 

=  15.0  X  10~3  microfarad  per  mile. 
b  =  2irf  X  15.0  X  10~3  X  10-6 
=  157  X  15.0  X  10-° 
=  2.355  X  10~6  mho  per  mile. 


SERIES-PARALLEL  CIRCUITS  419 

x  =  27T/  (741  loglo  1U^e12  +  80)  X  10-6 

=  157(741  X  2.584  +  80)  X  1Q-6 
=  0.3130  ohm  per  mile. 

z  =  \/r2  +  x2  =  V(0.1862)2  +  (0.3130)2 

=  0.3642  ohm  per  mile. 
0.3130 
tan  '•  =  OJ862  =  L681 

ez  =  +59.25  degrees  per  mile. 

z  =  0.36421+  59?250hm  per  mile. 


y  =  vV  +  V  =  V(O.OOO)2  +  (2.355  X  10~6)2 

=  2.355  X  10~6  mho  per  mile. 
2.355  X  10-6 


By  =  +90.0  degrees  per  mile. 
y  =  2.355  X  1Q-6|+90?0  ohm  per  mile. 
a  =  ^/^z  =  \/2.355  X  1Q-6  X  0.3642 

=  0.9261  X  10~3  per  mile. 
_  By  +  6Z  _  90?0  +  59°25 
6<x  ~        2  2 

=  74.63  degrees  per  mile. 
a  =  0.9261  X  10-3|+74°63  per  mile. 


fz  _      I      0.3642 
:  Vlf  ~  \2.355  X  1( 


.355  X  10-6. 
=  393.3  ohms. 
0,  -  Bv  =  59?25  -  90?00 
2  2 

=  - 15.38  degrees. 
393.3J- 15  ?38  ohms. 
1          1 


20      393.3 

=  0.002543  mho. 

By0  =  —  eZo  =  +15°38  degrees. 
yo  =  0.002543  +15°38  mho. 

sin  6a   =  0.9643 


420  PRINCIPLES  OF  ALTERNATING  CURRENTS 

cos  0«  =  0.2651 
sin  02o  =  -0.2651 
cos  0z0  =  0.9643 
sin  0y0  =  0.2651 
cos  By0  =  0.9643 
ai  =  a  cos  da  =  0.9261  X  10~3  X  0.2651 

=  0.2455  X  10~3  Ityperbolic  radians  per  mile. 
«2  =  a  sin  6a  =  0.9261  X  10~3  X  0.9643 

=  0.8931  X  10~3  radians  per  mile. 
aiL  =  0.2455  X  10~3  X  150 

=  0.03683  hyperbolic  radians. 
«2L  =  0.8931  X  10~3  X  150 
=  0.1340  radians. 

VR  (to  neutral)  =  110>Q°0  =  63,510  volts. 

V3 

r   /  N     30>°00  X  1000          1P__.n 

IR  (per  conductor)  3  x  63?51Q  x  Q  9Q  -  175.0  amperes. 

cos  BR  =  0.90         BR  =  25.84  degrees. 
Take  VR  as  the  axis  of  reference. 

VR  =  63,510  |0  f 0000 

I*  =  175.0|-25?84 
cosh  aiL  =  1.001 
sinh  aiL  =  0.03684 

1  Qf)° 

«2L  =  0.1340  X  - 

7T 

=  7.675  degrees, 
sin  a2L  =  0.1336 
cos  a2L  =  0.9912 
cosh  aL  =  cosh  a\L  cos  a^L  +  j  sinh  a\L  sin  aju 

=  1.001  X  0.9912  +  J0.03684  X  0.1336 

=  0.9922  +  J0.004920 

0.004920 


=  V(0.9922)2  +  (0.004920): 
=  0.9922 1 +0?284 


tan" 


0.9922 


SERIES-PARALLEL  CIRCUITS  421 

sinh  aL  =  sinh  a.\L  cos  a2L  +  j  cosh  ctiL  sin  a^L 
=  0.03684  X  0.9912  +j  1.001  X  0.1336 
=  0.03651  +j  0.1337 

0.1337 


V(0.03651)2  +  (0.1337); 


tan-1 


0.03651 


=  0.13861+ 74  °72 
V0  =  7fl  cosh  oL  -f-  HRzQ  sinh  oL 
=  63,510  lO°OQQ  X  0.9922 1 +0?28 

+  175.01 -25 °84  X  39&3  H15°38  x  0. 13861+ 74 ?72 
=  63,000  +0?28  +  9538i+33°50 


63,000(1.000  +  J0.00495)  +  9538(0.8338  +  J0.5521) 

70,950  +  J5578 

5578 


V  (70,950)2  +  (5578) : 


tan~ 


70,950 


=  71,170J+4°495  volts  to  neutral. 
VG  =  71,170  X  \/3  =  123,300  volts  between  conductors. 

.  123,300-110,000^ 

Line  regulation  for  the  given  load  =  -         Tin  000 —    ~ 

=  12.1  per  cent. 

The  no-load  voltage  at  the  load  with  123,300  volts  maintained 
at  the  generator  is 

VR  (no  load)  =  — -^= 

V  (cosh  ctiL  cos  azL)2  +  (sinh  a\L  sin  a^L)2 

123,300 
=  0.9922 
=  124,300  volts  between  conductors. 

The  percentage  rise  in  voltage  at  the  load-end  when  the  load 
is  removed,  the  generator  voltage  being  maintained  at  123,300 
volts,  is 


Io  =  IR  cosh  ^Jj  -f  Vxyo  sinh  «L 

=  175.0|-25°84  X  0.9922  1  +0°284 


+  63,510  10°000  X  0.002543  jf!5?38  x  0.1386  1  +74  °72 
=  l73.6'-25°56  +  22.38  1+90°10 
-  173.6  (0.9022  -  J0.4314)  +  22.38(-0.0017  +  j  1.000) 


422  PRINCIPLES  OF  ALTERNATING  CURRENTS 

=  156.6  -  J52.51 


=  \/(156.6)2  +  (52.51)2 


tan' 


-52.51 
156.6 


=  165.2  1  -  18  ?54  amperes. 

&G    =    By    —    01 

=  4°495  -  (-18°54) 
=  23.03  degrees. 

Power-factor  at  generator  =  cos  6G  =  cos  23  ?03 

=  0.9205 

PG  =  VGIG  cos  eG 

=  71,170  X  165.2  X  0.9205  X  10~3 
=  10,820  kilowatts  per  phase. 
By  complex  method 

PG  =  (70,950  X  156.6  -  5578  X  52.51)  X  10~3 
=  10,820  kilowatts  per  phase. 

10  000 
Efficiency  of  transmission  =  -  —^  X  100 


=  92.5  per  cent. 

When  the  generator  voltage  is  maintained  at  the  value  required 
to  give  110,000  volts  between  conductors  at  the  load,  the  load 
being  30,000  kilowatts  at  0.9  power-factor  (inductive),  the  no- 
load  charging  current  is 

7G  (charging)  =  IG    =  VR'yo  sinh  aL 
where  VR  is  the  voltage  at  the  load  on  open  circuit. 

VR' 
IG    =  ^r  (VRy0  sinh 

V  R 


The  magnitude  of  (FR#O  sinh  «L)  has  already  been  found  for  a 
voltage  at  the  load  of   VR  =  --  ^=—   volts  to  neutral.     (See 

calculation  of  the  generator  current  for  load  conditions.)  It  is 
the  second  term  of  the  second  member  in  the  expression  for  the 
generator  current.  The  charging  current  at  no  load  varies  di- 
rectly with  the  voltage. 

_  123,300 
"  110,000  > 

=  25.1  amperes. 


SERIES-PARALLEL  CIRCUITS  423 

Length  of  the  line  in  wave  lengths  is 


2  X  3.142 

where  X  is  the  wave  length  of  the  current  and  voltage  waves. 
Length  of  line,  neglecting  the  transmission  losses,  is 

T         Length  of  line  in  miles  X  frequency  v 

Lx  =    ,T  i     ., , ..  ,  .  . TJ —  -  X  wave  length 

Velocity  of  light  in  miles  per  sec. 

150  X  25 
"  186,000    X 
=  0.02016X 


INDEX 

Active  component,  of  current,  58;  of  voltage,  62. 

Active,  current,  59;  power,  60;  voltage,  63. 

Addition,  of  admittances,  216;  of  currents  or  voltages,  40,  43,  110;  of  imped- 
ances, 200. 

Admittance,  213,  243;  addition  of,  216;  complex  expression  for,  215;  polar 
expression  for,  217. 

Air -core  transformer,  induced  voltage  in  windings  of,  187;  vector  diagram 
of,  189. 

Alternating  current,  definition  of,  27;  ampere  value  of,  34,  82;  calculations 
based  on  sinusoidal  waves,  29;  measurement  of,  37,  79;  strength  of,  33. 

Alternating  voltage,  definition  of,  27;  generation  of,  31,  69,  245;  measure- 
ment of,  37,  79;  volt  value  of,  35,  84. 

Ampere  value,  of  an  alternating  current,  34,  82. 

Amplitude  factor,  106. 

Analysis,  of  non-sinusoidal  waves  into  fundamental  and  harmonics,  89,  92. 

Apparent  power,  55. 

Attenuation  constant,  413. 

Axis,  imaginary,  6;  real,  6. 

Balanced,  Y-connected  load  connected  across  a  three-phase  circuit  having 
balanced  voltages,  274,  277;  Y-  and  A-connected  loads  in  parallel,  277. 
Braking,  an  inductive  circuit,  127. 

Capacitance,  118;  effect  of,  119,  136,  137,  140,  142,  150;  of  two,  straight, 
parallel  conductors,  376;  of  two,  straight,  parallel  conductors,  per  1000 
feet  of  line,  385,  per  mile  of  line,  385;  between  earth  and  a  straight 
conductor  parallel  to  its  surface,  386;  between  earth  and  a  straight 
conductor  parallel  to  its  surface  per  1000  feet  of  conductor,  387,  per 
mile  of  conductor,  387;  of  a  balanced  three-phase  transmission  line  with 
conductors  at  corners  of  an  equilateral  triangle,  394,  per  mile,  396;  of  a 
three-phase  transmission  line  from  its  equivalent  equilateral  spacing, 
397. 

Charging  current,  of  a  transmission  line,  387,  391,  395. 

Circuit,  containing  constant  resistance  and  capacitance  in  series,  136,  137, 
140,  142,  149;  containing  constant  resistance  and  inductance  in  series, 
119,  121,  125,  127,  128,  134;  containing  constant  resistance,  inductance 
and  capacitance  in  series,  150,  151,  157,  161,  168;  coupled,  see  coupled 
circuits;  inductive,  breaking  of,  127;  filter,  229;  see  three-phase, 
four-phase  and  six-phase. 

Coefficient,  of  electromagnetic  coupling,  180;  of  leakage,  178;  of  leakage- 
induction,  183;  of  mutual-induction,  116,  174,  176,  180,  see  mutual 
inductance;  of  self-induction,  115,  180,  185. 

425 


426  INDEX 

Complex,  method,  3;  quantities,  3,  23,  135;  expression  for  admittance,  215; 
expression  for  impedance,  134. 

Components,  active,  energy,  or  power  of  a  current  or  voltage,  58,  61,  62; 
direct-phase,  of  an  unbalanced  three-phase  system,  338,  341,  345,  346; 
imaginary  of  a  vector,  6;  reactive,  wattless,  or  quadrature,  of  a  current 
or  voltage,  58,  61,  62;  real  of  a  vector,  6;  reverse-phase,  of  an  unbal- 
anced three-phase  system,  338,  341,  345,  346;  uniphase  components 
of  an  unbalanced  three-phase  system,  338,  341,  345,  346. 

Conductance,  213,  243. 

Connections,  delta,  255,  292;  four-phase,  262,  302;  mesh,  257,  262,  302,  306; 
six-phase,  305,  306;  star,  257,  262,  302,  306;  wye,  255,  289. 

Constants  of  a  transmission  line,  attenuation,  413;  propagation,  413; 
wave-length,  414. 

Copper,  relative  amounts  required  for  power  transmission  with  different 
numbers  of  phases,  320,  321,  322. 

Copper  loss,  in  an  unbalanced  three-phase  circuit,  352. 

Crest  factor,  106. 

Coupled  circuits,  differential  equation  for,  181;  effect  of  short-circuiting  one 
of  two  inductively  coupled  circuits  on  the  apparent  self-inductance  of 
the  other,  182;  leakage  coefficients  of,  178;  leakage  inductance  of,  183; 
magnetic  leakage  of,  178. 

Coupling,  close,  181;  coefficient  of  electromagnetic,  180. 

Current,  at  one  end  of  a  transmission  line  in  terms  of  line  constants  and 
current  and  voltage  at  other  end,  406,  410,  411;  active,  energy  or  power, 
58,  61;  active,  energy  or  power  component  of,  58,  61;  alternating,  defini- 
tion of,  27;  ampere  value  of,  34;  average  value  of,  35;  continuous,  defini- 
tion of,  27;  charging,  of  a  transmission  line,  387,  391,  395;  direct, 
definition  of,  27;  effective  value  of,  33,  34,  82;  instantaneous  value  of, 
27;  measurement  of  effective  or  root-mean-square  value  of,  37,  79; 
oscillating,  definition  of,  27;  positive  direction  of,  249;  pulsating, 
definition  of,  27;  ratio  of  average  and  effective  values  of,  for  sinusoidal 
waves,  36;  reactive,  wattless  or  quadrature,  58,  61;  reactive  wattless 
or  quadrature  components  of,  58,  61;  simple  harmonic,  definition  of,  29; 
strength  of,  33,  34,  82. 

Current,  non-sinusoidal,  analysis  of,  89,  92;  effective  oj  root-mean-square 
value  of,  82;  effective  or  root-mean-square  value  of,  from  polar  plot, 
88;  equivalent  sine  waves,  107;  measurement  of  effective  or  root-mean- 
square  value  of,  79;  representation  of  by  a  Fourier  Series,  72. 

Currents,  addition  of,  40,  43,  110;  polyphase,  generation  of,  245;  positive 
direction  of,  249;  relative  magnitudes  of  line  and  phase,  for  a  balanced 
three-phase  circuit,  258,  266,  291,  292;  for  a  balanced  four-phase  circuit, 
262,  266,  304,  for  a  balanced  six-phase  circuit,  266,  309;  for  a  balanced 
twelve-phase  circuit,  266,  for  a  balanced  n-phase  circuit,  264. 

Cycle,  27. 

Delta  connection,  255,  292;  equivalent  wye  connection,  280,  281,  282,  283, 
284,  294;  relative  magnitudes  of  line  and  phase  currents  and  voltages  and 


INDEX  427 

their  phase  relations,  258,  289,  292;  third  harmonic  short  circuited  in, 
292;  with  unbalanced  A-connected  load,  268;  with  single-phase  load,  261. 

Deviation  factor,  107. 

Direct,  current,  definition  of,  27;  voltage,  definition  of,  27. 

Direct-phase  components,  of  currents  and  voltages  of  an  unbalanced  three- 
phase  circuit,  338,  249,  252;  determination  of,  341,  345,  346. 

Double -subscript  notation,  248,  250. 

Effective,  reactance,  237;  resistance,  234;  value  of  an  alternating  current  or 
voltage,  33,  34,  82;  value  of  current  or  voltage,  measurement  of,  37,  79; 
value  of  current  or  voltage  from  a  polar  plot,  88. 

Electromotive  force,  see  voltage. 

Electrostatic  induction,  voltage  induced  by,  in  a  telephone  or  telegraph  line 
by  the  charges  on  the  conductors  of  an  adjacent  transmission  line,  399, 
401,  403. 

Energy,  component  of  current,  58;  component  of  voltage,  62;  of  electro- 
magnetic field,  116,  124;  of  electrostatic  field,  139. 

Equations,  vector,  solution  of,  12. 

Equivalent,  equilateral  spacing  of  a  three-phase  transmission  line,  397;  phase 
difference,  107;  resistance  and  reactance,  240;  sine  waves,  107;  Y-  and 
A-connected  circuits,  280,  281,  282,  284,  294. 

Equilateral  spacing,  equivalent,  of  a  three-phase  transmission  line,  397. 

Factor,  amplitude,  peak  or  crest,  106;  deviation,  107;  form,  37,  106. 

Filter  circuits,  229. 

Fisher -Hinnen,  method  of  analysing  a  periodic  non-sinusoidal  wave,  92. 

Flux,  mutual,  of  inductively  coupled  circuits,  186;  producing  mutual  induc- 
tance, 185;  producing  self -inductance,  185;  producing  leakage  induc- 
tance, 185. 

Form  factor,  37,  106. 

Four -phase,  alternators,  245,  262;  circuits,  connection  of,  262,  302;  har- 
monics in,  299,  301;  relative  magnitudes  of  line  and  phase  voltages  for, 
262,  266,  303. 

Fourier  Series,  for  a  non-sinusoidal  current  or  voltage,  72,  75;  determination 
of  constants  of,  89,  92;  for  rectangular  and  triangular  waves,  78;  for 
symmetrical  and  non-symmetrical  waves,  75. 

Frequency,  28,  247. 

Generation,  of  an  alternating  electromotive  force,  31,  69,  245. 

Harmonic,  73;  simple,  current  or  voltage,  29. 

Harmonics,  73;  effect  on  power,  85;  effect  on  power  factor,  86;  even,  effect 
of  on  wave  shape,  75;  in  balanced  three-phase  circuits,  287;  in  balanced 
four-phase  circuits,  299;  in  balanced  six-phase  circuits,  304;  third,  in 
neutral  connection  of  a  three-phase  Y-connected  circuit,  291;  third,  in 
neutral  connection  of  a  four-phase  star-connected  circuit,  304;  third, 
short-circuited  in  A-connected  circuit,  292;  third,  not  short-circuited 
in  four-phase  mesh-connected  circuit,  304. 

Henry,  116. 


428  INDEX 

Imaginary,  components,  6;  axis,  6. 

Impedance,  243;  of  a  circuit  containing  constant  resistance  and  self -induc- 
tance in  series,  134,  complex  expression  for,  134,  polar  expression  for, 
136;  of  a  circuit  containing  constant  resistance,  and  capacitance  in 
series,  148,  complex  expression  for,  149,  polar  expression  for,  150;  of 
a  circuit  containing  constant  resistance,  inductance  and  capacitance 
in  series,  168;  complex  expression  for,  169,  polar  expression  for,  170. 

Impedances,  equivalent  Y-  and  A-connected,  280,  281,  282,  283,  284;  in 
parallel,  211,  215,  243;  in  series,  196,  199,  243;  in  series-parallel,  224. 

Inductance,  of  a  single-phase  transmission  line,  359;  leakage,  183,  185; 
mutual,  116,  173,  174,  176,  180,  185. 

Induction,  electrostatic,  between  a  transmission  line  and  an  adjacent 
telephone  or  telegraph  line,  399,  401,  403;  mutual,  between  transmission 
lines  or  a  transmission  line  and  a  telephone  line,  367;  voltage,  electro- 
magnetic, induced  by,  in  a  telephone  line  by  a  three-phase  transmission 
line,  370,  371,  372;  voltage,  electrostatic,  induced  by,  in  a  telephone 
line  or  telegraph  line  by  the  charges  on  the  conductors  of  a  transmission 
line,  399. 

Instantaneous,  value  of  current  or  voltage,  27. 

j,  operator  which  rotates  a  vector  through  ninety  degrees,  4,  5,  10. 

Kirchhoff's  Laws,  271,  272. 

Leakage,  coefficient,  178;  flux,  178,  185;  inductance,  183;  magnetic,  178. 

Magnetic  leakage,  178. 

Magnitudes,  relative,  of  currents  and  voltages  in  polyphase  systems,  see 
delta,  wye,  four-phase,  six-phase,  twelve-phase  and  n-phase. 

Measurement,  of  current  and  voltage,  37,  79;  of  power  (active),  57,  79; 
of  reactive  power,  63. 

Mechanical  system,  having  constant  friction,  mass  and  elasticity,  170. 

Mesh  connection,  257,  262,  264,  302,  306. 

Mutual  flux,  of  inductively  coupled  circuits,  186. 

Mutual  induction,  116,  173,  174,  180,  185;  between  transmission  lines  or 
between  a  transmission  line  and  a  telephone  line,  348,  367;  of  two 
coupled  circuits  in  a  medium  of  constant  permeability,  176;  differential 
equation  for  a  coupled  circuit  having  constant  mutual  and  self-induc- 
tance, 176;  flux  corresponding  to,  185. 

Neutral,  third  harmonics  in,  for  a  three-phase  Y-connected  system,  291; 
for  a  four-phase  star-connected  system,  304. 

Non -sinusoidal  current  or  voltage  waves,  addition  and  subtraction  of,  110; 
effective  or  root- mean-square,  value  of,  82;  representation  of,  by  a 
Fourier  Series,  72. 

Non-sinusoidal  voltage,  impressed  on  a  circuit  containing  constant  imped- 
ances in  series  and  in  parallel,  232. 

N -phase  system,  power  in,  314;  relative  magnitudes  ii},  of  line  and  phase 
currents  for  sinusoidal  waves,  264. 


INDEX  429 

N -wattmeter  method,  for  measuring  power  in  an  n-phase  system,  326. 
(N-l) -wattmeter  method,  for  measuring  power  in  an  n-phase  system,  326. 

Operators,  (cos  a  ±  j  sin  a),  7,  8;  exponential,  15;  j,  4;  polar,  19;  powers  of, 
9,  24;  product  of,  9,  20;  reciprocal  of,  11;  roots  of,  9,  20,  24;  successive 
application  of,  9;  which  produce  uniform  angular  rotation,  11,  18. 

Oscillation,  155,  159,  162;  period  of,  for  a  circuit  containing  constant  resis- 
tance, inductance  and  capacitance  in  series,  208;  time  of,  for  a  circuit 
containing  constant  resistance,  inductance  and  capacitance  in  series, 
156,  159,  162;  of  a  mechanical  system,  having  constant  friction,  mass 
and  elasticity,  170. 

Parallel,  impedances  in,  211,  243;  resonance,  220. 

Peak  factor,  106. 

Period,  27;  of  oscillation  of  a  resonant  circuit  containing  constant  resistance, 
inductance  and  capacitance  in  series,  208. 

Periodic,  time,  27. 

Phase,  2,  30;  balancer,  350;  equivalent  difference  in,  for  non-sinusoidal 
waves,  107. 

Phase  order  of  harmonics,  in  a  balanced  three-phase  circuit,  287,  289;  in  a 
balanced  four-phase  circuit,  301;  in  a  balanced  six-phase  circuit,  307. 

Polar,  expression  for  admittance,  217;  expression  for  impedance,  136,  150, 
170;  form  of  operator  to  produce  rotation,  19. 

Polyphase  currents,  generation  of,  245;  see  mesh  and  star  connections,  three- 
phase,  four-phase,  six-phase,  n-phase,  delta  and  wye. 

Potential,  see  voltage. 

Power,  active,  60;  absorbed  and  delivered  by  a  circuit,  48;  average,  50; 
average,  when  current  and  voltage  waves  are  sinusoidal,  51,  52,  53; 
average,  when  current  and  voltage  waves  are  non-sinusoidal,  85; 
calculation  of,  from  complex  expressions  for  current  and  voltage,  66; 
component  of  current  or  voltage,  58,  62;  in  polyphase  circuit,  314; 
in  a  balanced  three-phase  circuit,  314;  in  an  unbalanced  three-phase 
circuit,  349;  instantaneous,  49;  measurement  of,  57,  79;  measurement 
of,  in  a  three-phase  circuit,  323,  324,  326 ;  measurement  of,  in  an  n-phase 
circuit,  326;  reactive,  60;  reactive,  calculation  of,  from  complex  expres- 
sions for  current  and  voltage,  66;  reactive,  measurement  of,  in  a  single- 
phase  circuit,  63;  reactive,  measurement  of,  hi  a  balanced  three-phase 
circuit,  336;  transfer  of,  by  mutual  induction  among  conductors  of  a 
three-phase  transmission  line,  365;  virtual,  55. 

Power  component,  of  current  or  voltage,  59,  63. 

Power  factor,  56,  61,  86;  of  balanced  polyphase,  circuit,  314,  317;  of  balanced 
three-phase  circuit,  316;  of  balanced  four-phase  circuit,  317;  of  unbal- 
anced polyphase  circuit,  317;  determination  of,  for  a  balanced  three- 
phase  circuit,  316,  333;  measurement  of,  58;  when  current  and  voltage 
waves  are  non-sinusoidal,  86,  316,  333. 

Powers,  of  operators,  9;  of  vectors,  24. 

Propagation  constant,  413. 


430  INDEX 

Quadrant,  116. 

Quadrature  component,  of  a  current,  58;  of  a  voltage,  62. 

Reactance,  average,  of  a  completely  transposed,  three-phase,  transmission 
line,  360,  363,  average,  of  a  completely  transposed,  three-phase,  trans- 
mission line  whose  conductors  are  at  the  corners  of  an  equilateral 
triangle,  363,  364;  average,  of  a  completely  transposed,  three-phase, 
transmission  line  whose  conductors  are  at  the  corners  of  an  isosceles 
triangle,  363;  average,  of  a  completely  transposed,  three-phase,  trans- 
mission line  whose  conductors  are  in  the  same  plane,  364;  capacitive, 
148,  168;  effective,  237,  243;  equivalent,  240;  function  of  current,  230;  in 
terms  of  conductance  and  susceptance,  216;  of  a  single-phase  trans- 
mission line,  355,  per  1000  feet,  359,  per  mile,  359. 

Reactive,  component  of  current,  58;  component  of  voltage,  62;  current,  59, 
61;  factor,  57;  power,  60;  power,  measurement  of,  in  a  single-phase 
circuit,  63;  power,  measurement  of,  in  a  balanced  three-phase  circuit, 
,336;  power,  measurement  of,  in  a  balanced  or  an  unbalanced  three- 
phase  circuit,  327a;  voltage,  63;  volt-amperes,  60 ;  volt-amperes,  measure- 
ment of,  63. 

Real,  axis,  6;  components,  6. 

Reference  point,  changing  the  position  of,  77. 

Relative  magnitudes  of  line  and  phase  currents  and  voltages,  in  a  balanced 
three-phase  system,  266,  289,  292,  294;  in  a  balanced  four-phase  system, 
266,  303;  in  a  balanced  six-phase  system,  266,  308,  310;  in  a  balanced 
twelve-phase  system,  266. 

Resistance,  effective,  234,  243;  equivalent,  240;  function  of  current,  230; 
in  terms  of  conductance  and  susceptance,  216. 

Residuals,  or  uniphase  components  in  an  unbalanced  three-phase  system, 
337,  see  uniphase  components. 

Resonance,  parallel,  220;  series,  204. 

Reverse-phase  components,  in  an  unbalanced  three-phase  circuit,  338,  249, 
252;  determination  of,  341,  345,  346. 

Root-mean-square  value,  of  an  alternating  current  or  voltage,  33,  35; 
measurement  of,  37,  79;  relation  between  root-mean-square  and  average 
values,  36. 

Roots,  of  operators,  9,  20. 

Secohm,  116. 

Self-inductance,  115,  180,  185;  effect  of,  117,  119,  121,  125,  127,  128,  150; 
function  of  current  when  magnetic  material  is  present,  125. 

Self-induction,  coefficient  of,  see  self-inductance. 

Series  circuit,  containing  constant  resistance,  and  inductance,  119,  121,  125, 
127,  128,  134;  containing  constant  resistance  and  capacitance,  136,  137, 
140,  142,  149;  containing  constant  resistance,  inductance  and  capaci- 
tance, 150,  151,  157,  161,  168. 

Series-parallel,  impedances,  224,  circuits,  405. 

Series  resonance,  204. 

Simple  harmonic,  current  or  voltage,  29. 

Six-phase  circuits,  harmonics  in,  304,  307;  relative  magnitude  of  line  and 
phase  voltages  for,  266,  308. 


INDEX  431 

Skin  effect,  236. 

Star  connection,  257,  262,  302,  306. 

Strength,  of  current,  33. 

Subscript,  double,  notation  for,  248,  250. 

Subtraction,  of  non-sinusoidal  waves,  110. 

Susceptance,  213,  243. 

Summary,  of  conditions  in  series  and  in  parallel  circuits,  243. 

Third  harmonics,  in  three-phase  circuits,  287,  289,  292,  297;  in  four-phase 
circuits,  299;  in  six-phase  circuits,  304;  in  neutral  connection  of  aY- 
connected  three-phase  circuit,  291;  in  neutral  connection  of  a  star- 
connected  circuit,  304;  in  a  mesh-connected  circuit,  304,  309;  short- 
circuited  in  a  three-phase  A-connected  circuit,  292. 

Third -harmonic  current,  in  neutral  connection  of  a  Y-connected  three- 
phase  circuit,  291;  in  a  mesh-connected  four-phase  circuit,  304;  in 
mesh-connected  six-phase  circuit,  309;  in  neutral  of  four-phase  star- 
connected  circuit,  304;  short  circuited  in  A-connected  three-phase 
circuit,  292. 

Three-phase,  alternators,  advantage  of,  248;  circuits,  delta  connection  of, 
292;  circuits,  harmonics  in,  287,  see  harmonics  and  third  harmonics; 
circuits,  wye  connection  of,  289;  circuits,  relative  magnitude  of  line  and 
phase  currents  and  voltages  for,  266,  289,  294. 

Three-phase  connection,  see  three-phase  circuits,  delta  and  wye  connection; 
of  generator,  advantages  of,  248. 

Three-wattmeter  method,  for  measuring  power  in  a  three-phase  circuit, 
323,  324. 

Time,  periodic,  27. 

Transfer  of  power,  by  mutual  induction,  among  conductors  of  a  three-phase 
transmission  line,  365. 

Transformer,  see  air-core  transformer. 

Transmission  line,  calculation  of  performance  of,  418;  capacitance  of,  see 
capacitance;  direct  and  reflected  waves  in,  411;  equations  for,  406,  410, 
411,  415,  416;  inductance  of,  see  inductance,  induction  and  mutual- 
induction;  reactance  of,  see  reactance. 

Two -wattmeter,  method  for  measuring  power  in  a  three-phase  circuit,  326; 
relative  magnitude  of  the  wattmeter  readings  for  balanced  load,  330. 

Uniphase,  components  of  an  unbalanced  three-phase  circuit,  337,  338,  341, 
249,  252. 

Vector,  algebra,  2;  diagrams  of  air-core  transformer,  189;  equations,  solu- 
tion of,  12;  oscillating,  25;  representation  by  use  of  operator  j,  4; 
representing  a  simple-harmonic  current  or  voltage,  39;  representing  a 
voltage  rise  or  fall,  64",  249;  roots  and  powers  of,  20,  24;  rotating,  which 
decrease  in  magnitude  with  time,  26. 

Vectors,  product  and  ratio  of,  20;  addition  of,  12,  40,  43. 

Virtual  power,  55. 

Volt,  value  of  an  alternating  voltage,  35,  82. 


432  INDEX 

Voltage,  active,  62;  active  component  of,  62;  alternating,  definition  of,  27; 
analysis  of,  into  its  fundamental  and  harmonics,  89,  92;  at  one  end  of 
a  transmission  line  in  terms  of  line  constants  and  voltage  and  current 
at  other  end,  406,  410,  411;  continuous,  definition  of,  27;  direct,  defini- 
tion of,  27;  effective  value  of,  35,  82;  effective  value  of,  from  a  polar 
plot,  88;  energy  component  of,  62;  equivalent  sine  wave,  107;  fallx  48, 
49,  50,  64,  249;  generation  of,  31,  69,  245;  induced  in  alternator,  31, 
69,  245;  induced  by  electromagnetic  induction  in  a  telephone  circuit 
by  a  three-phase  transmission  line,  370,  371,  372;  induced  by  electro- 
static induction  in  a  telephone  line  by  the  charges  on  the  conductors 
of  a  transmission  line,  399;  induced  in  the  windings  of  an  air-core  trans- 
former, 187;  instantaneous,  definition  of,  27;  measurement  of  effective 
or  root-mean-square  value  of,  37,  79;  non-sinusoidal,  representation  of, 
by  Fourier  Series,  72;  pulsating,  definition  of,  27;  quadrature  component 
of,  62;  reactive  component  of,  62;  rise,  48,  49,  50,  64,  249;  simple 
harmonic,  29;  wattless  component  of,  62. 

Voltages,  equivalent  wye  and  delta,  294;  relative  magnitude  of  line  and 
phase,  for  three-phase  circuits,  258,  266,  290,  294;  relative  magnitude 
of  line  and  phase,  for  four-phase  circuits,  266,  303;  relative  magnitude 
of  line  and  phase,  for  six-phase  circuits,  266,  308. 

Volt-amperes,  55,  61. 

Wattless  component,  of  current,  58;  of  voltage,  62. 

Wattmeter,  57,  79;  two-wattmeter  method,  326,  330;  three-wattmeter 
method,  323,  324;  n-wattmeter  method,  323,  326;  (n-1) -wattmeter 
method,  326. 

Wave,  analysis  of,  89,  92;  equivalent  sine,  107,  108;  non-sinusoidal,  29,  75, 
76;  non-sinusoidal,  representation  of  by  a  Fourier  Series,  72;  shape  or 
form,  28;  sinusoidal,  29. 

Wave  form,  28;  determination  of,  81;  of  alternators,  31,  69;  similar,  77. 

Wave  length,  for  transmission  line,  steady  state,  414;  constant,  414. 

Wye  connection,  255,  289;  equivalent  delta  connection,  280,  281,  282,  294; 
relative  magnitude  of  line  and  phase  voltages,  258,  266,  289;  third  har- 
monics in  neutral  of,  291. 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below, 


JUN1 


• 

INTER  LIBRARY 
LOAN 

IE  MONTH  AFTER  RECE1P 

<'2     1985 


LD  21-95w^ll,'50(2877sl6)476 


re 


L 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


